第四章 不定积分

1\. 利用导数验证下列等式：

(1)

$$ \int\frac{1}{\sqrt{x^{2}+1}}\mathrm{d}x=\ln\left(x+\sqrt{x^{2}+1}\right)+C; $$

(2) $ \int \frac{1}{x^2 \sqrt{x^2 - 1}} \, dx = \frac{\sqrt{x^2 - 1}}{x} + C $;

(3) $ \int \frac{2x}{\left(x^{2}+1\right)\left(x+1\right)^{2}} \mathrm{d}x = \arctan x + \frac{1}{x+1} + C $;

(4) $ \int \sec x \, dx = \ln |\tan x + \sec x| + C $;

(5) $ \int x\cos xdx = x\sin x + \cos x + C $;

(6) $ \int e^{x}\sin x dx = \frac{1}{2}e^{x}(\sin x - \cos x) + C. $

解 (1) $ \frac{\mathrm{d}}{\mathrm{d}x}\left[\ln\left(x+\sqrt{x^{2}+1}\right)+C\right]=\frac{1}{x+\sqrt{x^{2}+1}}\cdot\left(1+\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{1}{\sqrt{x^{2}+1}} $

(2)

$$ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\sqrt{x^{2}-1}}{x}+C\right)=\frac{\frac{x}{\sqrt{x^{2}-1}}\cdot x-\sqrt{x^{2}-1}}{x^{2}}=\frac{1}{x^{2}\sqrt{x^{2}-1}}. $$

$$ \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\arctan x+\frac{1}{x+1}+C\bigg)=\frac{1}{x^{2}+1}-\frac{1}{\left(x+1\right)^{2}}=\frac{2x}{\left(x^{2}+1\right)\left(x+1\right)^{2}}. $$

$$ \frac{\mathrm{d}}{\mathrm{d}x}(\ln|\tan x+\sec x|+C)=\frac{1}{\tan x+\sec x}\cdot(\sec^{2}x+\sec x\tan x)=\sec x. $$

(5)

$$ \frac{\mathrm{d}}{\mathrm{d}x}(x\sin x+\cos x+C)=\sin x+x\cos x-\sin x=x\cos x. $$

(6)

$$ \begin{aligned}\frac{\mathrm{d}}{\mathrm{d}x}\bigg[\frac{1}{2}\mathrm{e}^{x}(\sin x-\cos x)+C\bigg]&=\frac{1}{2}\mathrm{e}^{x}(\sin x-\cos x)+\frac{1}{2}\mathrm{e}^{x}(\cos x+\sin x)\\&=\mathrm{e}^{x}\sin x.\end{aligned} $$

1. 求下列不定积分：

(1)

$$ \int\frac{\mathrm{d}x}{x^{2}}; $$

(2)

$$ \int x\sqrt{x}\mathrm{d}x; $$

(3) $ \int\frac{dx}{\sqrt{x}} $;

(4) $ \int x^{2}\sqrt[3]{x}dx $;

(5) $ \int\frac{dx}{x^{2}\sqrt{x}}; $

(6) $ \int \sqrt[m]{x^n} \, dx $;

(7) $ \int 5x^{3} \, dx $;

(8) $ \int (x^{2} - 3x + 2) \, dx $;

(9) $ \int \frac{\mathrm{d}h}{\sqrt{2gh}} $ (g 是常数);

(10) $ \int (x^{2} + 1)^{2} \, \mathrm{d}x $;

(11) $ \int (\sqrt{x} + 1) (\sqrt{x^3} - 1) \, \mathrm{d}x $;

(12) $ \int \frac{(1 - x)^2}{\sqrt{x}} \, dx $;

(13) $ \int\left(2e^{x}+\frac{3}{x}\right)dx $;

(14) $ \int\left(\frac{3}{1+x^{2}}-\frac{2}{\sqrt{1-x^{2}}}\right)dx; $

(15) $ \int \mathrm{e}^{x}\left(1 - \frac{\mathrm{e}^{-x}}{\sqrt{x}}\right) \mathrm{d}x $;

(16) $ \int 3^{x} e^{x} dx $;

(17) $ \int \frac{2 \cdot 3^x - 5 \cdot 2^x}{3^x} \, dx $;

(18) $ \int \sec x (\sec x - \tan x) \, dx $;

(19) $ \int\cos^{2}\frac{x}{2}dx $;

(20) $ \int \frac{dx}{1 + \cos 2x} $;

(21) $ \int \frac{\cos 2x}{\cos x - \sin x} \, dx $;

(22) $ \int \frac{\cos 2x}{\cos^2 x \sin^2 x} \, dx $;

(23) $ \int \cot^{2} x \, dx $;

(24) $ \int \cos \theta (\tan \theta + \sec \theta) \, \mathrm{d}\theta $;

(25) $ \int\frac{x^{2}}{x^{2}+1}dx $;

(26) $ \int\frac{3x^{4}+2x^{2}}{x^{2}+1}dx. $

解 (1) $ \int \frac{dx}{x^{2}} = \int x^{-2} dx = \frac{1}{-2 + 1} x^{-2 + 1} + C = -\frac{1}{x} + C. $

(2) $ \int x \sqrt{x} \, dx = \int x^{\frac{3}{2}} \, dx = \frac{1}{\frac{3}{2} + 1} x^{\frac{3}{2} + 1} + C = \frac{2}{5} x^{\frac{5}{2}} + C. $

(3) $ \int \frac{dx}{\sqrt{x}} = \int x^{-\frac{1}{2}} dx = \frac{1}{-\frac{1}{2} + 1} x^{-\frac{1}{2} + 1} + C = 2 \sqrt{x} + C. $

(4) $ \int x^{2}\sqrt[3]{x}\mathrm{d}x = \int x^{\frac{7}{3}}\mathrm{d}x = \frac{1}{\frac{7}{3}+1}x^{\frac{7}{3}+1} + C = \frac{3}{10}x^{\frac{10}{3}} + C. $

(5) $ \int \frac{dx}{x^{2} \sqrt{x}} = \int x^{-\frac{5}{2}} dx = \frac{1}{-\frac{5}{2} + 1} x^{-\frac{5}{2} + 1} + C = -\frac{2}{3} x^{-\frac{3}{2}} + C. $

(6)

$$ \int\sqrt[m]{x^{n}}\mathrm{d}x=\frac{1}{\frac{n}{m}+1}x^{\frac{n}{m}+1}+C=\frac{m}{m+n}x^{\frac{m+n}{m}}+C. $$

(7)

$$ \int5x^{3}\mathrm{d}x=\frac{5}{3+1}x^{3+1}+C=\frac{5}{4}x^{4}+C. $$

(8)

$$ \int\left(x^{2}-3x+2\right)\mathrm{d}x=\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x+2\int\mathrm{d}x=\frac{x^{3}}{3}-\frac{3}{2}x^{2}+2x+C. $$

(9)

$$ \int\frac{\mathrm{d}h}{\sqrt{2gh}}=\frac{1}{\sqrt{2g}}\int h^{-\frac{1}{2}}\mathrm{d}h=\frac{1}{\sqrt{2g}}\times2\sqrt{h}+C=\sqrt{\frac{2h}{g}}+C. $$

(10)

$$ \begin{aligned}\int\left(x^{2}+1\right)^{2}\mathrm{d}x&=\int\left(x^{4}+2x^{2}+1\right)\mathrm{d}x=\int x^{4}\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int\mathrm{d}x\\&=\frac{x^{5}}{5}+\frac{2}{3}x^{3}+x+C.\end{aligned} $$

(11)

$$ \begin{aligned}\int\left(\sqrt{x}+1\right)\left(\sqrt{x^{3}}-1\right)\mathrm{d}x&=\int\left(x^{2}+x^{\frac{3}{2}}-x^{\frac{1}{2}}-1\right)\mathrm{d}x\\&=\int x^{2}\mathrm{d}x+\int x^{\frac{3}{2}}\mathrm{d}x-\int x^{\frac{1}{2}}\mathrm{d}x-\int\mathrm{d}x\\&=\frac{x^{3}}{3}+\frac{2}{5}x^{\frac{5}{2}}-\frac{2}{3}x^{\frac{3}{2}}-x+C.\end{aligned} $$

(12)

$$ \begin{align*}\int\frac{\left(1-x\right)^{2}}{\sqrt{x}}\mathrm{d}x&=\int\left(x^{\frac{3}{2}}-2x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right)\mathrm{d}x\\&=\int x^{\frac{3}{2}}\mathrm{d}x-2\int x^{\frac{1}{2}}\mathrm{d}x+\int x^{-\frac{1}{2}}\mathrm{d}x\\&=\frac{2}{5}x^{\frac{5}{2}}-\frac{4}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C.\end{align*} $$

(13)

$$ \int\left(2\mathrm{e}^{x}+\frac{3}{x}\right)\mathrm{d}x=2\int\mathrm{e}^{x}\mathrm{d}x+3\int\frac{\mathrm{d}x}{x}=2\mathrm{e}^{x}+3\ln|x|+C. $$

(14)

$$ \begin{aligned}\int\left(\frac{3}{1+x^{2}}-\frac{2}{\sqrt{1-x^{2}}}\right)\mathrm{d}x&=3\int\frac{\mathrm{d}x}{1+x^{2}}-2\int\frac{\mathrm{d}x}{\sqrt{1-x^{2}}}\\&=3\arctan x-2\arcsin x+C.\end{aligned} $$

(15)

$$ \int\mathrm{e}^{x}\left(1-\frac{\mathrm{e}^{-x}}{\sqrt{x}}\right)\mathrm{d}x=\int\mathrm{e}^{x}\mathrm{d}x-\int x^{-\frac{1}{2}}\mathrm{d}x=\mathrm{e}^{x}-2x^{\frac{1}{2}}+C. $$

(16)

$$ \int3^{x}\mathrm{e}^{x}\mathrm{d}x=\int\left(3\mathrm{e}\right)^{x}\mathrm{d}x=\frac{\left(3\mathrm{e}\right)^{x}}{\ln\left(3\mathrm{e}\right)}+C=\frac{3^{x}\mathrm{e}^{x}}{\ln3+1}+C. $$

(17)

$$ \begin{aligned}\int\frac{2\cdot3^{x}-5\cdot2^{x}}{3^{x}}\mathrm{d}x&=2\int\mathrm{d}x-5\int\left(\frac{2}{3}\right)^{x}\mathrm{d}x=2x-\frac{5}{\ln\frac{2}{3}}\bigg(\frac{2}{3}\bigg)^{x}+C\\&=2x-\frac{5}{\ln2-\ln3}\bigg(\frac{2}{3}\bigg)^{x}+C.\end{aligned} $$

(18)

$$ \begin{aligned}\int\sec x\left(\sec x-\tan x\right)\mathrm{d}x&=\int\sec^{2}x\mathrm{d}x-\int\sec x\tan x\mathrm{d}x\\&=\tan x-\sec x+C.\end{aligned} $$

(19)

$$ \int\cos^{2}\frac{x}{2}\mathrm{d}x=\int\frac{1+\cos x}{2}\mathrm{d}x=\frac{x+\sin x}{2}+C. $$

(20)

$$ \int\frac{\mathrm{d}x}{1+\cos2x}=\int\frac{\sec^{2}x}{2}\mathrm{d}x=\frac{\tan x}{2}+C. $$

(21)

$$ \int\frac{\cos2x}{\cos x-\sin x}\mathrm{d}x=\int\frac{\cos^{2}x-\sin^{2}x}{\cos x-\sin x}\mathrm{d}x=\sin x-\cos x+C. $$

(22)

$$ \begin{aligned}\int\frac{\cos2x}{\cos^{2}x\sin^{2}x}\mathrm{d}x&=\int\frac{\cos^{2}x-\sin^{2}x}{\cos^{2}x\sin^{2}x}\mathrm{d}x=\int\left(\csc^{2}x-\sec^{2}x\right)\mathrm{d}x\\&=\int\csc^{2}x\mathrm{d}x-\int\sec^{2}x\mathrm{d}x=-\left(\cot x+\tan x\right)+C.\end{aligned} $$

(23)

$$ \int\cot^{2}x\mathrm{d}x=\int\csc^{2}x\mathrm{d}x-\int\mathrm{d}x=-\cot x-x+C. $$

(24)

$$ \int\cos\theta(\tan\theta+\sec\theta)\mathrm{d}\theta=\int\sin\theta\mathrm{d}\theta+\int\mathrm{d}\theta=-\cos\theta+\theta+C. $$

(25)

$$ \int\frac{x^{2}}{x^{2}+1}\mathrm{d}x=\int\mathrm{d}x-\int\frac{1}{x^{2}+1}\mathrm{d}x=x-\arctan x+C. $$

(26)

$$ \int\frac{3x^{4}+2x^{2}}{x^{2}+1}\mathrm{d}x=\int3x^{2}\mathrm{d}x-\int\mathrm{d}x+\int\frac{1}{x^{2}+1}\mathrm{d}x=x^{3}-x+\arctan x+C. $$

1. 含有未知函数的导数的方程称为微分方程，例如方程 $ \frac{dy}{dx}=f(x) $，其中 $ \frac{dy}{dx} $ 为未知函数的导数， $ f(x) $ 为已知函数。如果将函数 $ y=\varphi(x) $ 代入微分方程，使微分方程成为恒等式，那么函数 $ y=\varphi(x) $ 就称为这个微分方程的解。求下列微分方程满足所给条件的解：

(1) $ \frac{dy}{dx} = (x - 2)^2 $, $ y|_{x=2} = 0 $;

(2) $ \frac{d^{2}x}{dt^{2}}=\frac{2}{t^{3}},\left.\frac{dx}{dt}\right|_{t=1}=1,x|_{t=1}=1. $

解（1）

$$ y=\int\left(x-2\right)^{2}\mathrm{d}x=\frac{1}{3}\left(x-2\right)^{3}+C, $$

由 $ y|_{x=2}=0 $ , 得 C=0 ，于是所求的解为 $ y=\frac{1}{3}(x-2)^{3} $

(2)

$$ \frac{\mathrm{d}x}{\mathrm{d}t}=\int\frac{2}{t^{3}}\mathrm{d}t=-\frac{1}{t^{2}}+C_{1}, $$

由 $ \left.\frac{dx}{dt}\right|_{t=1}=1 $，得 $ C_{1}=2 $，故 $ \frac{dx}{dt}=-\frac{1}{t^{2}}+2 $，

$$ x~=~\int\left(-\frac{1}{t^{2}}+2\right)\mathrm{d}t=\frac{1}{t}+2t+C_{2}, $$

由 $ x\big|_{t=1}=1 $ ，得 C_{2}=-2，于是所求的解为 $ x=\frac{1}{t}+2t-2 $

1. 汽车以 $ 20 \, m/s $ 的速度行驶，刹车后匀减速行驶了 $ 50 \, m $ 停住，求刹车加速度。可执行下列步骤：

（1）求微分方程 $ \frac{d^{2}s}{dt^{2}}=-k $ 满足条件 $ \left.\frac{ds}{dt}\right|_{t=0}=20 $ 及 $ s|_{t=0}=0 $ 的解；

（2）求使 $ \frac{ds}{dt}=0 $的t值；

（3）求使 s=50 的 k 值.

解（1）

$$ \frac{\mathrm{d}s}{\mathrm{d}t}=\int-k\mathrm{d}t=-\ k t+C_{1}, $$

由 $ \frac{ds}{dt}\bigg|_{t=0}=20 $，得 $ C_{1}=20 $，故

$$ \frac{\mathrm{d}s}{\mathrm{d}t}=-k t+20, $$

$$ s=\int\left(-k t+20\right)\mathrm{d}t=-\frac{1}{2}k t^{2}+20t+C_{2}, $$

由 $ \left.s\right|_{t=0}=0 $ ，得 $ C_{2}=0 $ ，于是所求的解为

$$ s=-\frac{1}{2}kt^{2}+20t. $$

（2）令 $ \frac{ds}{dt}=0 $，解得 $ t=\frac{20}{k} $

（3）根据题意，当 $ t=\frac{20}{k} $时，s=50，即

$$ -\frac{1}{2}k\left(\frac{20}{k}\right)^{2}+\frac{400}{k}=50, $$

解得 k=4，即得刹车加速度为 $ -4 \, m/s^{2} $

1. 一曲线通过点 $ (e^{2},3) $，且在任一点处的切线的斜率等于该点横坐标的倒数，求该曲线的方程.

解 设曲线方程为 $ y = f(x) $，则点 $ (x, y) $ 处的切线斜率为 $ f'(x) $，由条件得

$$ f^{\prime}\left(x\right)=\frac{1}{x}, $$

因此 $ f(x) $ 为 $ \frac{1}{x} $ 的一个原函数，故有 $ f(x) = \int \frac{1}{x} \mathrm{d}x = \ln |x| + C $.

又，根据条件曲线过点 $ (e^{2},3) $，有 $ f(e^{2})=3 $解得C=1，即得所求曲线方程为

$$ y=\ln x+1. $$

1. 一物体由静止开始运动，经 t 秒后的速度是 $ 3t^{2} $ (m/s)，问

（1）在3秒后物体离开出发点的距离是多少？

（2）物体走完360m需要多少时间？

解 （1）设此物体自原点沿横轴正向由静止开始运动，位移函数为 $ s = s(t) $，则

$$ s\left(t\right)=\int v\left(t\right)\mathrm{d}t=\int3t^{2}\mathrm{d}t=t^{3}+C, $$

由假设可知 $ s(0)=0 $ ，故 $ s(t)=t^{3} $ ，于是所求距离为 $ s(3)=27 $ (m).

（2）由 $ t^{3}=360 $，得 $ t=\sqrt[3]{360}\approx7.11\ (s) $

1. 证明函数 $ \arcsin(2x-1) $, $ \arccos(1-2x) $ 和 $ 2\arctan\sqrt{\frac{x}{1-x}} $ 都是 $ \frac{1}{\sqrt{x-x^{2}}} $ 的原函数.

$$ \left[\arcsin\left(2x-1\right)\right]^{\prime}=\frac{1}{\sqrt{1-\left(2x-1\right)^{2}}}\cdot2=\frac{1}{\sqrt{x-x^{2}}}, $$

$$ \left[\arccos(1-2x)\right]^{\prime}=-\frac{1}{\sqrt{1-(1-2x)^{2}}}\cdot(-2)=\frac{1}{\sqrt{x-x^{2}}}, $$

$$ \left(2\arctan\sqrt{\frac{x}{1-x}}\right)^{\prime}=2\frac{1}{1+\frac{x}{1-x}}\cdot\frac{1}{2}\sqrt{\frac{1-x}{x}}\cdot\frac{1}{\left(1-x\right)^{2}}=\frac{1}{\sqrt{x-x^{2}}}. $$

故结论成立.

习题4-2

换元积分法

在下列各式等号右端的空白处填入适当的系数，使等式成立（例如：

$ \frac{1}{4}\mathrm{d}(4x+7) $ ：

(1) $ \mathrm{d}x = \underline{\mathrm{d}}(ax) $ ；

(2) $ \mathrm{d}x = \underline{\mathrm{d}}(7x-3) $ ；

(3) $ x\mathrm{d}x = \underline{\mathrm{d}}(x^{2}) $ ；

(4) $ x\mathrm{d}x = \underline{\mathrm{d}}(5x^{2}) $ ；

(5) $ x\mathrm{d}x = \underline{\mathrm{d}}(1-x^{2}) $ ；

(6) $ x^{3}\mathrm{d}x = \underline{\mathrm{d}}(3x^{4}-2) $ ；

(7) $ \mathrm{e}^{2x}\mathrm{d}x = \underline{\mathrm{d}}(\mathrm{e}^{2x}) $ ；

(8) $ \mathrm{e}^{-\frac{1}{2}}\mathrm{d}x = \underline{\mathrm{d}}(1+\mathrm{e}^{-\frac{1}{2}}) $ ；

(9) $ \sin\frac{3}{2}x\mathrm{d}x = \underline{\mathrm{d}}(\cos\frac{3}{2}x) $ ；

(10) $ \frac{\mathrm{d}x}{x} = \underline{\mathrm{d}}(5\ln|x|) $ ；

(11) $ \frac{\mathrm{d}x}{x} = \underline{\mathrm{d}}(3-5\ln|x|) $ ；

(12) $ \frac{\mathrm{d}x}{1+9x^{2}} = \underline{\mathrm{d}}(\arctan3x) $ ；

(13) $ \frac{\mathrm{d}x}{\sqrt{1-x^{2}}} = \underline{\mathrm{d}}(1-\arcsin x) $ ；

(14) $ \frac{x\mathrm{d}x}{\sqrt{1-x^{2}}} = \underline{\mathrm{d}}(\sqrt{1-x^{2}}) $ 。

解 (1) $ \frac{1}{a} $ ；

(2) $ \frac{1}{7} $ ；

(3) $ \frac{1}{2} $ ；

(4) $ \frac{1}{10} $ ；

(5) $ -\frac{1}{2} $ ；

(6) $ \frac{1}{12} $ ；

(7) $ \frac{1}{2} $ ；

(8) -2；

(9) $ -\frac{2}{3} $ ；

(10) $ \frac{1}{5} $ ；

$$ \mathrm{d}x= $$

(11) $ -\frac{1}{5} $; (12) $ \frac{1}{3} $; (13) -1; (14) -1.

1. 求下列不定积分（其中 a、b、 $ \omega $、 $ \varphi $ 均为常数）：

(1) $ \int e^{5t} dt $;

(2) $ \int (3 - 2x)^{3} \, \mathrm{d}x $;

(3) $ \int\frac{\mathrm{d}x}{1-2x}; $

(4) $ \int \frac{\mathrm{d}x}{\sqrt[3]{2-3x}}; $

(5) $ \int (\sin ax - e^{\frac{x}{b}}) \, dx $;

(6) $ \int \frac{\sin\sqrt{t}}{\sqrt{t}}dt $;

(7) $ \int x e^{-x^{2}} \, dx $;

(8) $ \int x\cos(x^{2})\,dx $;

(9) $ \int \frac{x}{\sqrt{2 - 3x^{2}}} \, dx $;

(10) $ \int \frac{3x^{3}}{1 - x^{4}} \mathrm{d}x $;

(11) $ \int \frac{x + 1}{x^{2} + 2x + 5} \, dx $;

(12) $ \int \cos^{2}(\omega t + \varphi) \sin(\omega t + \varphi) \, dt $;

(13) $ \int \frac{\sin x}{\cos^{3}x} \, dx $;

(14) $ \int \frac{\sin x + \cos x}{\sqrt[3]{\sin x - \cos x}} \, dx $;

(15) $ \int \tan^{10} x \cdot \sec^{2} x \, dx $;

(16) $ \int \frac{dx}{x \ln x \ln \ln x} $;

(17) $ \int \frac{dx}{(\arcsin x)^2 \sqrt{1 - x^2}}; $

(18) $ \int \frac{10^{2\arccos x}}{\sqrt{1-x^2}} \, dx $;

(19) $ \int \tan \sqrt{1 + x^2} \cdot \frac{x \, dx}{\sqrt{1 + x^2}} $;

(20) $ \int \frac{\arctan \sqrt{x}}{\sqrt{x}(1 + x)} \, dx $;

(21) $ \int \frac{1 + \ln x}{(x \ln x)^2} \, dx $;

(22) $ \int \frac{dx}{\sin x \cos x} $;

(23) $ \int \frac{\ln \tan x}{\cos x \sin x} \, dx $;

(24) $ \int \cos^{3} x \, dx $;

(25) $ \int \cos^{2}\left(\omega t + \varphi\right) \, dt $;

(26) $ \int \sin 2x \cos 3x \, dx $;

(27) $ \int \cos x \cos \frac{x}{2} \, dx $;

(28) $ \int \sin 5x \sin 7x \, dx $;

(29) $ \int \tan^{3} x \sec x \, dx $;

(30) $ \int \frac{dx}{e^x + e^{-x}}; $

(31) $ \int\frac{1-x}{\sqrt{9-4x^{2}}}dx $;

(32) $ \int \frac{x^{3}}{9 + x^{2}} \mathrm{d}x $;

(33) $ \int \frac{dx}{2x^{2}-1} $;

(34) $ \int \frac{dx}{(x+1)(x-2)} $;

(35)

$$ \int\frac{x}{x^{2}-x-2}\mathrm{d}x; $$

(36) $ \int \frac{x^{2} \, \mathrm{d}x}{\sqrt{a^{2} - x^{2}}} (a > 0) $;

(37) $ \int \frac{dx}{x\sqrt{x^{2}-1}} $;

(38) $ \int \frac{dx}{\sqrt{(x^{2}+1)^{3}}} $;

(39) $ \int \frac{\sqrt{x^{2}-9}}{x} \mathrm{d}x $;

(40) $ \int \frac{dx}{1 + \sqrt{2x}}; $

(41) $ \int \frac{dx}{1 + \sqrt{1 - x^2}}; $

(42) $ \int \frac{dx}{x + \sqrt{1 - x^2}}; $

(43) $ \int \frac{x - 1}{x^{2} + 2x + 3} \, dx $;

(44) $ \int \frac{x^{3} + 1}{\left(x^{2} + 1\right)^{2}} \, dx $.

解（1）令 u=5t，由第一类换元法得

$$ \int\mathrm{e}^{5t}\mathrm{d}t=\frac{1}{5}\int\mathrm{e}^{u}\mathrm{d}u=\frac{1}{5}\mathrm{e}^{u}+C=\frac{1}{5}\mathrm{e}^{5t}+C. $$

（2）令u=3-2x，由第一类换元法得

$$ \int\left(3-2x\right)^{3}\mathrm{d}x=-\frac{1}{2}\int u^{3}\mathrm{d}u=-\frac{u^{4}}{8}+C=-\frac{\left(3-2x\right)^{4}}{8}+C. $$

（3）令u=1-2x，由第一类换元法得

$$ \int\frac{\mathrm{d}x}{1-2x}=-\frac{1}{2}\int\frac{\mathrm{d}u}{u}=-\frac{1}{2}\ln\left|u\right|+C=-\frac{1}{2}\ln\left|1-2x\right|+C. $$

$$ \begin{aligned}(4)\int\frac{\mathrm{d}x}{\sqrt[3]{2-3x}}&=\int-\frac{1}{3}(2-3x)^{-\frac{1}{3}}\mathrm{d}(2-3x)\\&=-\frac{1}{3}\cdot\frac{3}{2}(2-3x)^{\frac{2}{3}}+C=-\frac{1}{2}(2-3x)^{\frac{2}{3}}+C.\end{aligned} $$

(5)

$$ \begin{aligned}5)\int(\sin ax-\mathrm{e}^{\frac{x}{b}})\mathrm{d}x&=\int\sin ax\mathrm{d}x-\int\mathrm{e}^{\frac{x}{b}}\mathrm{d}x\\&=\int\frac{1}{a}\sin ax\mathrm{d}(ax)-\int b\mathrm{e}^{\frac{x}{b}}\mathrm{d}\left(\frac{x}{b}\right)\\&=\frac{1}{a}(-\cos ax)-b\mathrm{e}^{\frac{x}{b}}+C=-\frac{\cos ax}{a}-b\mathrm{e}^{\frac{x}{b}}+C.\end{aligned} $$

(6) $ \int \frac{\sin\sqrt{t}}{\sqrt{t}}dt = \int 2\sin\sqrt{t}d\sqrt{t} = -2\cos\sqrt{t} + C. $

(7)

$$ \int x\mathrm{e}^{-x^{2}}\mathrm{d}x=-\frac{1}{2}\int\mathrm{e}^{-x^{2}}\mathrm{d}\left(-x^{2}\right)=-\frac{1}{2}\mathrm{e}^{-x^{2}}+C. $$

$$ \int x\cos\left(x^{2}\right)\mathrm{d}x=\frac{1}{2}\int\cos\left(x^{2}\right)\mathrm{d}\left(x^{2}\right)=\frac{1}{2}\sin\left(x^{2}\right)+C. $$

(9)

$$ \int\frac{x}{\sqrt{2-3x^{2}}}\mathrm{d}x=-\frac{1}{6}\int\left(2-3x^{2}\right)^{-\frac{1}{2}}\mathrm{d}\left(2-3x^{2}\right) $$

$$ =-\frac{1}{6}\cdot2\left(2-3x^{2}\right)^{\frac{1}{2}}+C=-\frac{\sqrt{2-3x^{2}}}{3}+C. $$

(10)

$$ \int\frac{3x^{3}}{1-x^{4}}\mathrm{d}x=-\frac{3}{4}\int\frac{1}{1-x^{4}}\mathrm{d}(1-x^{4})=-\frac{3}{4}\ln|1-x^{4}|+C. $$

(11)

$$ \int\frac{x+1}{x^{2}+2x+5}\mathrm{d}x=\frac{1}{2}\int\frac{\mathrm{d}(x^{2}+2x+5)}{x^{2}+2x+5}=\frac{1}{2}\ln(x^{2}+2x+5)+C. $$

（12）

$$ \begin{aligned}\int\cos^{2}\left(\omega t+\varphi\right)\sin\left(\omega t+\varphi\right)\mathrm{d}t&=-\frac{1}{\omega}\int\cos^{2}\left(\omega t+\varphi\right)\mathrm{d}\left[\cos\left(\omega t+\varphi\right)\right]\\&=-\frac{1}{3\omega}\cos^{3}\left(\omega t+\varphi\right)+C.\end{aligned} $$

(13)

$$ \int\frac{\sin x}{\cos^{3}x}\mathrm{d}x=-\int\frac{1}{\cos^{3}x}\mathrm{d}(\cos x)=\frac{1}{2\cos^{2}x}+C. $$

（14）

$$ \int\frac{\sin x+\cos x}{\sqrt[3]{\sin x-\cos x}}\mathrm{d}x=\int\frac{\mathrm{d}(\sin x-\cos x)}{\sqrt[3]{\sin x-\cos x}}=\frac{3}{2}(\sin x-\cos x)^{\frac{2}{3}}+C. $$

（15）

$$ \int\tan^{10}x\cdot\sec^{2}x\mathrm{d}x=\int\tan^{10}x\mathrm{d}\left(\tan x\right)=\frac{1}{11}\tan^{11}x+C. $$

(16)

$$ \int\frac{\mathrm{d}x}{x\ln x\ln\ln x}=\int\frac{\mathrm{d}(\ln x)}{\ln x\ln\ln x}=\int\frac{\mathrm{d}(\ln\ln x)}{\ln\ln x}=\ln|\ln\ln x|+C. $$

（17）

$$ \int\frac{\mathrm{d}x}{\left(\arcsin x\right)^{2}\sqrt{1-x^{2}}}=\int\frac{\mathrm{d}\left(\arcsin x\right)}{\left(\arcsin x\right)^{2}}=-\frac{1}{\arcsin x}+C. $$

(18)

$$ \int\frac{10^{2\arccos x}}{\sqrt{1-x^{2}}}\mathrm{d}x=\int-10^{2\arccos x}\mathrm{d}(\arccos x)=-\frac{10^{2\arccos x}}{2\ln10}+C. $$

(19)

$$ \begin{align*}\int\tan\sqrt{1+x^{2}}\cdot\frac{x\mathrm{d}x}{\sqrt{1+x^{2}}}&=\frac{1}{2}\int\tan\sqrt{1+x^{2}}\cdot\frac{\mathrm{d}(1+x^{2})}{\sqrt{1+x^{2}}}\\&=\int\tan\sqrt{1+x^{2}}\mathrm{d}(\sqrt{1+x^{2}})\\&=-\ln\left|\cos\sqrt{1+x^{2}}\right|+C.\end{align*} $$

(20)

$$ \begin{aligned}\int\frac{\arctan\sqrt{x}}{\sqrt{x}\left(1+x\right)}\mathrm{d}x&=\int\frac{2\arctan\sqrt{x}}{1+x}\mathrm{d}\sqrt{x}=\int2\arctan\sqrt{x}\mathrm{d}\left(\arctan\sqrt{x}\right)\\&=\left(\arctan\sqrt{x}\right)^{2}+C.\end{aligned} $$

(21)

$$ \int\frac{1+\ln x}{\left(x\ln x\right)^{2}}\mathrm{d}x=\int\frac{\mathrm{d}\left(x\ln x\right)}{\left(x\ln x\right)^{2}}=-\frac{1}{x\ln x}+C. $$

(22)

$$ \int\frac{\mathrm{d}x}{\sin x\cos x}=\int\csc2x\mathrm{d}(2x)=\ln|\csc2x-\cot2x|+C=\ln|\tan x|+C. $$

(23)

$$ \begin{aligned}\int\frac{\ln\tan x}{\cos x\sin x}\mathrm{d}x&=\int\frac{\ln\tan x}{\tan x}\mathrm{d}(\tan x)=\int\ln\tan x\mathrm{d}(\ln\tan x)\\&=\frac{(\ln\tan x)^{2}}{2}+C.\end{aligned} $$

(24)

$$ \int\cos^{3}x\mathrm{d}x=\int\left(1-\sin^{2}x\right)\mathrm{d}(\sin x)=\sin x-\frac{1}{3}\sin^{3}x+C. $$

(25)

$$ \int\cos^{2}\left(\omega t+\varphi\right)dt=\int\frac{\cos2\left(\omega t+\varphi\right)+1}{2}dt=\frac{\sin2\left(\omega t+\varphi\right)}{4\omega}+\frac{t}{2}+C. $$

(26)

$$ \int\sin2x\cos3x\mathrm{d}x=\int\frac{1}{2}(\sin5x-\sin x)\mathrm{d}x=-\frac{1}{10}\cos5x+\frac{1}{2}\cos x+C. $$

(27)

$$ \begin{aligned}\int\cos x\cos\frac{x}{2}\mathrm{d}x&=\int\frac{1}{2}\bigg(\cos\frac{3}{2}x+\cos\frac{1}{2}x\bigg)\mathrm{d}x\\&=\frac{1}{3}\sin\frac{3}{2}x+\sin\frac{1}{2}x+C.\end{aligned} $$

(28)

$$ \begin{align*}\int\sin5x\sin7x\mathrm{d}x&=\int-\frac{1}{2}(\cos12x-\cos2x)\mathrm{d}x\\&=-\frac{1}{24}\sin12x+\frac{1}{4}\sin2x+C.\end{align*} $$

(29)

$$ \int\tan^{3}x\sec x\mathrm{d}x=\int\left(\sec^{2}x-1\right)\mathrm{d}(\sec x)=\frac{1}{3}\sec^{3}x-\sec x+C. $$

(30)

$$ \int\frac{\mathrm{d}x}{\mathrm{e}^{x}+\mathrm{e}^{-x}}=\int\frac{\mathrm{e}^{x}\mathrm{d}x}{\mathrm{e}^{2x}+1}=\int\frac{\mathrm{d}\left(\mathrm{e}^{x}\right)}{\mathrm{e}^{2x}+1}=\arctan\left(\mathrm{e}^{x}\right)+C. $$

(31)

$$ \begin{array}{l}\displaystyle\int\frac{1-x}{\sqrt{9-4x^{2}}}\mathrm{d}x~=\frac{1}{2}\int\frac{\mathrm{d}\left(\frac{2x}{3}\right)}{\sqrt{1-\left(\frac{2x}{3}\right)^{2}}}+\frac{1}{8}\int\frac{\mathrm{d}(9-4x^{2})}{\sqrt{9-4x^{2}}}\\ \\ \displaystyle\quad=\frac{\arcsin\frac{2x}{3}}{2}+\frac{\sqrt{9-4x^{2}}}{4}+C.\end{array} $$

(32)

$$ \int\frac{x^{3}}{9+x^{2}}\mathrm{d}x=\int x\mathrm{d}x-\frac{9}{2}\int\frac{\mathrm{d}(9+x^{2})}{9+x^{2}}=\frac{x^{2}}{2}-\frac{9}{2}\ln(9+x^{2})+C. $$

(33)

$$ \int\frac{\mathrm{d}x}{2x^{2}-1}=\frac{1}{2}\int\left(\frac{1}{\sqrt{2}x-1}-\frac{1}{\sqrt{2}x+1}\right)\mathrm{d}x=\frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}x-1}{\sqrt{2}x+1}\right|+C. $$

(34)

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\left(x+1\right)\left(x-2\right)}&=\int\frac{1}{3}\bigg(\frac{1}{x-2}-\frac{1}{x+1}\bigg)\mathrm{d}x\\&=\frac{1}{3}\ln\left|\frac{x-2}{x+1}\right|+C.\end{aligned} $$

(35)

$$ \begin{aligned}\int\frac{x}{x^{2}-x-2}\mathrm{d}x&=\int\frac{x}{\left(x-2\right)\left(x+1\right)}\mathrm{d}x=\int\frac{1}{3}\bigg(\frac{2}{x-2}+\frac{1}{x+1}\bigg)\mathrm{d}x\\&=\frac{2}{3}\ln\left|x-2\right|+\frac{1}{3}\ln\left|x+1\right|+C.\end{aligned} $$

（36）设 $ x = a \sin u \left( -\frac{\pi}{2} < u < \frac{\pi}{2} \right) $，则 $ \sqrt{a^{2} - x^{2}} = a \cos u $， $ dx = a \cos u \mathrm{d}u $，于是

$$ \begin{align*}\int\frac{x^{2}\mathrm{d}x}{\sqrt{a^{2}-x^{2}}}&=\int a^{2}\sin^{2}u\mathrm{d}u=a^{2}\int\frac{1-\cos2u}{2}\mathrm{d}u\\&=\frac{a^{2}}{2}\bigg(u-\frac{\sin2u}{2}\bigg)+C\\&=\frac{a^{2}}{2}\arcsin\frac{x}{a}-\frac{x\sqrt{a^{2}-x^{2}}}{2}+C.\end{align*} $$

（37）当x>1时，

$$ \begin{aligned}\int\frac{\mathrm{d}x}{x\sqrt{x^{2}-1}}&\xlongequal{x=\frac{1}{t}}-\int\frac{\mathrm{d}t}{\sqrt{1-t^{2}}}=-\arcsin t+C\\&=-\arcsin\frac{1}{x}+C,\end{aligned} $$

当x < -1时，

$$ \int\frac{\mathrm{d}x}{x\sqrt{x^{2}-1}}\xlongequal{x=\frac{1}{t}}\int\frac{\mathrm{d}t}{\sqrt{1-t^{2}}}=\arcsin t+C=\arcsin\frac{1}{x}+C, $$

故在 $ (-∞, -1) $或 $ (1, +∞) $内，有

$$ \int\frac{\mathrm{d}x}{x\sqrt{x^{2}-1}}=-\arcsin\frac{1}{\left|x\right|}+C. $$

（38）设 $ x = \tan u \left( -\frac{\pi}{2} < u < \frac{\pi}{2} \right) $，则 $ \sqrt{x^{2} + 1} = \sec u $， $ \mathrm{d}x = \sec^{2} u \, \mathrm{d}u $，于是

$$ \int\frac{\mathrm{d}x}{\sqrt{\left(x^{2}+1\right)^{3}}}=\int\cos u\mathrm{d}u=\sin u+C=\frac{x}{\sqrt{1+x^{2}}}+C. $$

（39）当 x > 3 时，令 $ x = 3\sec u $ ( $ 0 \leq u < \frac{\pi}{2} $)，

$$ \begin{align*}\int\frac{\sqrt{x^{2}-9}}{x}\mathrm{d}x&=\int3\tan^{2}u\mathrm{d}u=3\int\left(\sec^{2}u-1\right)\mathrm{d}u=3\tan u-3u+C\\&=\sqrt{x^{2}-9}-3\arccos\frac{3}{x}+C;\end{align*} $$

当 $x < -3$ 时，令 $x = 3\sec u\left(\frac{\pi}{2} < u \leq \pi\right)$，

$$ \begin{array}{r l}{\displaystyle\int\frac{\sqrt{x^{2}-9}}{x}\mathrm{d}x}&{=-\displaystyle\int3\tan^{2}u\mathrm{d}u=-3\displaystyle\int\left(\sec^{2}u-1\right)\mathrm{d}u=-3\tan u+3u+C^{\prime}}\\ &{=\displaystyle\sqrt{x^{2}-9}+3\operatorname{a r c c o s}\frac{3}{x}+C^{\prime}}\\ &{=\displaystyle\sqrt{x^{2}-9}-3\operatorname{a r c c o s}\frac{3}{-x}+C^{\prime}+3\pi,}\end{array} $$

故可统一写作 $ \int\frac{\sqrt{x^{2}-9}}{x}dx=\sqrt{x^{2}-9}-3\arccos\frac{3}{|x|}+C. $

$$ \int\frac{\mathrm{d}x}{1+\sqrt{2x}}\xlongequal{x=\frac{u^{2}}{2}}\int\frac{u\mathrm{d}u}{1+u}=u-\ln(1+u)+C=\sqrt{2x}-\ln(1+\sqrt{2x})+C. $$

（41）令 $ x = \sin t \left( -\frac{\pi}{2} < t < \frac{\pi}{2} \right) $，则 $ \sqrt{1 - x^{2}} = \cos t $， $ dx = \cos t dt $，于是

$$ \begin{align*}\int\frac{\mathrm{d}x}{1+\sqrt{1-x^{2}}}&=\int\frac{\cos t}{1+\cos t}\mathrm{d}t=\int\frac{2\cos^{2}\frac{t}{2}-1}{2\cos^{2}\frac{t}{2}}\mathrm{d}t=t-\tan\frac{t}{2}+C\\&=t-\frac{\sin t}{1+\cos t}+C=\arcsin x-\frac{x}{1+\sqrt{1-x^{2}}}+C.\end{align*} $$

（42）设 $ x = \sin t \left(-\frac{\pi}{4} < t < \frac{\pi}{2}\right) $，则 $ \sqrt{1 - x^{2}} = \cos t $， $ dx = \cos t dt $，于是

$$ \int\frac{\mathrm{d}x}{x+\sqrt{1-x^{2}}}=\int\frac{\cos t\mathrm{d}t}{\sin t+\cos t}, $$

记 $ I_{1}=\int\frac{\cos t\,dt}{\sin t+\cos t}, I_{2}=\int\frac{\sin t\,dt}{\sin t+\cos t} $，利用

$$ I_{1}+I_{2}=\int\mathrm{d}t=t+C, $$

$$ I_{1}-I_{2}=\int\frac{\cos t-\sin t}{\sin t+\cos t}\mathrm{d}t=\int\frac{\mathrm{d}(\sin t+\cos t)}{\sin t+\cos t}=\ln|\sin t+\cos t|+C, $$

求得

$$ I_{1}=\int\frac{\cos t\mathrm{d}t}{\sin t+\cos t}=\frac{1}{2}(t+\ln|\sin t+\cos t|)+C, $$

即求得在 $ \left(-\frac{\sqrt{2}}{2},1\right) $内，有

$$ \int\frac{\mathrm{d}x}{x+\sqrt{1-x^{2}}}=\frac{1}{2}(\arcsin x+\ln|x+\sqrt{1-x^{2}}|)+C； $$

再设 $ x = \sin t \left(-\frac{\pi}{2} < t < -\frac{\pi}{4}\right) $，重复上面的过程，可得在 $ (-1, -\frac{\sqrt{2}}{2}) $ 内有与上面不定积分形式相同的结果。从而在 $ (-1, -\frac{\sqrt{2}}{2}) $ 或 $ \left(-\frac{\sqrt{2}}{2}, 1\right) $ 内，有

$$ \int\frac{\mathrm{d}x}{x+\sqrt{1-x^{2}}}=\frac{1}{2}(\arcsin x+\ln|x+\sqrt{1-x^{2}}|)+C. $$

(43)

$$ \begin{align*}\int\frac{x-1}{x^{2}+2x+3}\mathrm{d}x&=\int\frac{x+1-2}{\left(x+1\right)^{2}+2}\mathrm{d}x=\frac{1}{2}\int\frac{\mathrm{d}\left[\left(x+1\right)^{2}+2\right]}{\left(x+1\right)^{2}+2}-\sqrt{2}\int\frac{\mathrm{d}\left(\frac{x+1}{\sqrt{2}}\right)}{\left(\frac{x+1}{\sqrt{2}}\right)^{2}+1}\\&=\frac{1}{2}\ln\left(x^{2}+2x+3\right)-\sqrt{2}\arctan\frac{x+1}{\sqrt{2}}+C.\end{align*} $$

（44）设 $ x = \tan t \left(-\frac{\pi}{2} < t < \frac{\pi}{2}\right) $，则 $ x^{2} + 1 = \sec^{2} t $， $ \mathrm{d}x = \sec^{2} t \mathrm{d}t $，于是

$$ \begin{aligned}\int\frac{x^{3}+1}{\left(x^{2}+1\right)^{2}}\mathrm{d}x&=\int\frac{\tan^{3}t+1}{\sec^{2}t}\mathrm{d}t\\&=\int\frac{\cos^{2}t-1}{\cos t}\mathrm{d}(\cos t)+\int\frac{1+\cos2t}{2}\mathrm{d}t\\&=\frac{1}{2}\cos^{2}t-\ln\cos t+\frac{t}{2}+\frac{1}{4}\sin2t+C\\&=\frac{1}{2}\cos^{2}t-\ln\cos t+\frac{t}{2}+\frac{1}{2}\sin t\cos t+C.\\ \end{aligned} $$

按 $ \tan t = x $ 作辅助三角形（图 4-1），便有

$$ \cos t=\frac{1}{\sqrt{1+x^{2}}},\quad\sin t=\frac{x}{\sqrt{1+x^{2}}}, $$

于是

$$ \int\frac{x^{3}+1}{\left(x^{2}+1\right)^{2}}\mathrm{d}x=\frac{1+x}{2\left(1+x^{2}\right)}+\frac{1}{2}\ln\left(1+x^{2}\right)+\frac{1}{2}\arctan x+C. $$

习题4-3

分部积分法

求下列不定积分：

1. $ \int x \sin x \, dx $.

1. $ \int \arcsin x \, dx $.

$$ \int\ln x\mathrm{d}x. $$

$$ \int x\mathrm{e}^{-x}\mathrm{d}x. $$

5.

$$ \int x^{2}\ln x\mathrm{d}x. $$

$$ \int\mathrm{e}^{-x}\cos x\mathrm{d}x. $$

7.

$$ \int\mathrm{e}^{-2x}\sin\frac{x}{2}\mathrm{d}x. $$

8.

$$ \int x\cos\frac{x}{2}\mathrm{d}x. $$

9.

$$ \int x^{2}\arctan x\mathrm{d}x. $$

10.

$$ \int x\tan^{2}x\mathrm{d}x. $$

11.

$$ \int x^{2}\cos x\mathrm{d}x. $$

12.

$$ \int t\mathbf{e}^{-2t}\mathrm{d}t. $$

13.

$$ \int\ln^{2}x\mathrm{d}x. $$

14.

$$ \int x\sin x\cos x\mathrm{d}x. $$

15.

$$ \int x^{2}\cos^{2}\frac{x}{2}\mathrm{d}x. $$

16.

$$ \int x\ln(x-1)\mathrm{d}x. $$

17.

$$ \int\left(x^{2}-1\right)\sin2x\mathrm{d}x. $$

18.

$$ \int\frac{\ln^{3}x}{x^{2}}\mathrm{d}x. $$

19.

$$ \int\mathrm{e}^{\sqrt{x}}\mathrm{d}x. $$

20.

$$ \int\cos\ln x\mathrm{d}x. $$

21.

$$ \int\left(\arcsin x\right)^{2}\mathrm{d}x. $$

22.

$$ \int\mathrm{e}^{x}\sin^{2}x\mathrm{d}x. $$

23.

$$ \int x\ln^{2}x\mathrm{d}x. $$

24.

$$ \int\mathrm{e}^{\sqrt{3x+9}}\mathrm{d}x. $$

$$ \begin{aligned}1.\ \int x\sin x\mathrm{d}x&=-\int x\mathrm{d}(\cos x)=-x\cos x+\int\cos x\mathrm{d}x\\ &=-x\cos x+\sin x+C.\end{aligned} $$

2.

$$ \int\ln x\mathrm{d}x=x\ln x-\int x\cdot\frac{1}{x}\mathrm{d}x=x\ln x-x+C. $$

$$ 3.\ \int\arcsin x\mathrm{d}x=x\arcsin x-\int x\cdot\frac{1}{\sqrt{1-x^{2}}}\mathrm{d}x=x\arcsin x+\sqrt{1-x^{2}}+C. $$

$$ 4.\int x\mathrm{e}^{-x}\mathrm{d}x=-\int x\mathrm{d}\mathrm{e}^{-x}=-x\mathrm{e}^{-x}+\int\mathrm{e}^{-x}\mathrm{d}x=-x\mathrm{e}^{-x}-\mathrm{e}^{-x}+C. $$

$$ 5.\int x^{2}\ln x\mathrm{d}x=\frac{1}{3}\int\ln x\mathrm{d}(x^{3})=\frac{x^{3}\ln x}{3}-\frac{1}{3}\int x^{3}\cdot\frac{1}{x}\mathrm{d}x=\frac{x^{3}\ln x}{3}-\frac{x^{3}}{9}+C. $$

6.

$$ \begin{array}{r l}{\int\mathrm{e}^{-x}\cos x\mathrm{d}x}&{=-\displaystyle\int\cos x\mathrm{d}\left(\mathrm{e}^{-x}\right)=-\mathrm{e}^{-x}\cos x+\displaystyle\int\mathrm{e}^{-x}\left(-\sin x\right)\mathrm{d}x}\\ &{}\\ &{=-\mathrm{e}^{-x}\cos x+\displaystyle\int\sin x\mathrm{d}\left(\mathrm{e}^{-x}\right)}\\ &{}\\ &{=-\mathrm{e}^{-x}\cos x+\mathrm{e}^{-x}\sin x-\displaystyle\int\mathrm{e}^{-x}\cos x\mathrm{d}x,}\end{array} $$

故有

$$ \int\mathrm{e}^{-x}\cos x\mathrm{d}x=\frac{\mathrm{e}^{-x}(\sin x-\cos x)}{2}+C. $$

$$ 7.\ \int\mathrm{e}^{-2x}\sin\frac{x}{2}\mathrm{d}x=-\frac{1}{2}\int\sin\frac{x}{2}\mathrm{d}(\mathrm{e}^{-2x}) $$

$$ \begin{aligned}&=-\frac{1}{2}\mathrm{e}^{-2x}\sin\frac{x}{2}+\frac{1}{2}\int\mathrm{e}^{-2x}\cdot\frac{1}{2}\cos\frac{x}{2}\mathrm{d}x\\&=\quad-\frac{1}{2}\mathrm{e}^{-2x}\sin\frac{x}{2}-\frac{1}{8}\int\cos\frac{x}{2}\mathrm{d}(\mathrm{e}^{-2x})\\&=-\frac{1}{2}\mathrm{e}^{-2x}\sin\frac{x}{2}-\frac{1}{8}\mathrm{e}^{-2x}\cos\frac{x}{2}+\frac{1}{8}\int\mathrm{e}^{-2x}\cdot\left(-\frac{1}{2}\sin\frac{x}{2}\right)\mathrm{d}x\\&=-\frac{1}{8}\Big(4\sin\frac{x}{2}+\cos\frac{x}{2}\Big)\mathrm{e}^{-2x}-\frac{1}{16}\int\mathrm{e}^{-2x}\sin\frac{x}{2}\mathrm{d}x,\end{aligned} $$

故

$$ \int\mathrm{e}^{-2x}\sin\frac{x}{2}\mathrm{d}x=-\frac{2}{17}\bigg(4\sin\frac{x}{2}+\cos\frac{x}{2}\bigg)\mathrm{e}^{-2x}+C. $$

8.

$$ \begin{align*}\int x\cos\frac{x}{2}\mathrm{d}x&=2\int x\mathrm{d}\bigg(\sin\frac{x}{2}\bigg)=2x\sin\frac{x}{2}-2\int\sin\frac{x}{2}\mathrm{d}x\\&=2x\sin\frac{x}{2}+4\cos\frac{x}{2}+C.\end{align*} $$

9.

$$ \begin{array}{r l}{\int x^{2}\arctan x\mathrm{d}x}&{=\frac{1}{3}\int\arctan x\mathrm{d}\left(x^{3}\right)=\frac{1}{3}x^{3}\arctan x-\frac{1}{3}\int\frac{x^{3}}{1+x^{2}}\mathrm{d}x}\\ &{}\\ &{=\frac{1}{3}x^{3}\arctan x-\frac{1}{3}\int\left(x-\frac{x}{1+x^{2}}\right)\mathrm{d}x}\\ &{}\\ &{=\frac{1}{3}x^{3}\arctan x-\frac{1}{6}x^{2}+\frac{1}{6}\ln(1+x^{2})+C.}\end{array} $$

10.

$$ \begin{align*}\int x\tan^{2}x\mathrm{d}x&=\int x(\sec^{2}x-1)\mathrm{d}x=\int x\mathrm{d}(\tan x)-\frac{x^{2}}{2}\\&=x\tan x+\ln|\cos x|-\frac{x^{2}}{2}+C.\end{align*} $$

11.

$$ \begin{aligned}\int x^{2}\cos x\mathrm{d}x&=\int x^{2}\mathrm{d}(\sin x)=x^{2}\sin x-\int2x\sin x\mathrm{d}x\\&=x^{2}\sin x+\int2x\mathrm{d}(\cos x)\\&=x^{2}\sin x+2x\cos x-\int2\cos x\mathrm{d}x\\&=x^{2}\sin x+2x\cos x-2\sin x+C.\end{aligned} $$

12.

$$ \begin{align*}\int t\mathrm{e}^{-2t}\mathrm{d}t&=-\frac{1}{2}\int t\mathrm{d}(\mathrm{e}^{-2t})=-\frac{1}{2}t\mathrm{e}^{-2t}+\frac{1}{2}\int\mathrm{e}^{-2t}\mathrm{d}t\\&=-\frac{1}{2}t\mathrm{e}^{-2t}-\frac{1}{4}\mathrm{e}^{-2t}+C.\end{align*} $$

$$ \begin{aligned}\int\ln^{2}x\mathrm{d}x&=x\ln^{2}x-\int2\ln x\mathrm{d}x=x\ln^{2}x-2x\ln x+\int2\mathrm{d}x\\&=x\ln^{2}x-2x\ln x+2x+C.\end{aligned} $$

14.

$$ \begin{aligned}\int x\sin x\cos x\mathrm{d}x&=\int-\frac{x}{4}\mathrm{d}(\cos2x)=-\frac{x\cos2x}{4}+\frac{1}{4}\int\cos2x\mathrm{d}x\\&=-\frac{x\cos2x}{4}+\frac{\sin2x}{8}+C.\end{aligned} $$

15.

$$ \begin{aligned}\int x^{2}\cos^{2}\frac{x}{2}\mathrm{d}x&=\frac{1}{2}\int x^{2}\left(1+\cos x\right)\mathrm{d}x=\frac{1}{6}x^{3}+\frac{1}{2}\int x^{2}\mathrm{d}(\sin x)\\&=\frac{1}{6}x^{3}+\frac{1}{2}x^{2}\sin x-\int x\sin x\mathrm{d}x\\&=\frac{1}{6}x^{3}+\frac{1}{2}x^{2}\sin x+\int x\mathrm{d}(\cos x)\\&=\frac{1}{6}x^{3}+\frac{1}{2}x^{2}\sin x+x\cos x-\int\cos x\mathrm{d}x\\&=\frac{1}{6}x^{3}+\frac{1}{2}x^{2}\sin x+x\cos x-\sin x+C.\end{aligned} $$

16.

$$ \begin{align*}\delta.\ \int x\ln\left(x-1\right)\mathrm{d}x&=\frac{1}{2}\int\ln\left(x-1\right)\mathrm{d}\left(x^{2}-1\right)\\&=\frac{1}{2}\left(x^{2}-1\right)\ln\left(x-1\right)-\frac{1}{2}\int\left(x+1\right)\mathrm{d}x\\&=\frac{1}{2}\left(x^{2}-1\right)\ln\left(x-1\right)-\frac{1}{4}x^{2}-\frac{1}{2}x+C.\end{align*} $$

17.

$$ \begin{aligned}7.\int\left(x^{2}-1\right)\sin2x\mathrm{d}x&=-\frac{1}{2}\int\left(x^{2}-1\right)\mathrm{d}(\cos2x)\\&=-\frac{1}{2}\left(x^{2}-1\right)\cos2x+\int x\cos2x\mathrm{d}x\\&=-\frac{1}{2}\left(x^{2}-1\right)\cos2x+\frac{1}{2}\int x\mathrm{d}(\sin2x)\\&=-\frac{1}{2}\left(x^{2}-1\right)\cos2x+\frac{1}{2}x\sin2x-\frac{1}{2}\int\sin2x\mathrm{d}x\\&=-\frac{1}{2}\left(x^{2}-\frac{3}{2}\right)\cos2x+\frac{1}{2}x\sin2x+C.\end{aligned} $$

18.

$$ \begin{align*}\int\frac{\ln^{3}x}{x^{2}}\mathrm{d}x&=\int-\ln^{3}x\mathrm{d}\Big(\frac{1}{x}\Big)=-\frac{\ln^{3}x}{x}-3\int\ln^{2}x\mathrm{d}\Big(\frac{1}{x}\Big)\\&=-\frac{\ln^{3}x}{x}-3\Big[\frac{\ln^{2}x}{x}+2\int\ln x\mathrm{d}\Big(\frac{1}{x}\Big)\Big]\\&=-\frac{\ln^{3}x+3\ln^{2}x+6\ln x+6}{x}+C.\end{align*} $$

19.

$$ \begin{aligned}\int\mathrm{e}^{\sqrt[3]{x}}\mathrm{d}x&\xlongequal{x=u^{3}}\int3u^{2}\mathrm{e}^{u}\mathrm{d}u=\int3u^{2}\mathrm{d}(\mathrm{e}^{u})=3u^{2}\mathrm{e}^{u}-\int6u\mathrm{d}(\mathrm{e}^{u})\\ &=\left(3u^{2}-6u+6\right)\mathrm{e}^{u}+C=3\mathrm{e}^{\sqrt[3]{x}}\left(x^{2/3}-2x^{1/3}+2\right)+C.\end{aligned} $$

1. $ \int \cos \ln x \, dx \xlongequal{x = e^{u}} \int e^{u} \cos u \, du $,

而

$$ \begin{align*}\int\mathrm{e}^{u}\cos u\mathrm{d}u&=\int\cos u\mathrm{d}(\mathrm{e}^{u})=\mathrm{e}^{u}\cos u+\int\mathrm{e}^{u}\sin u\mathrm{d}u\\&=\mathrm{e}^{u}\cos u+\int\sin u\mathrm{d}(\mathrm{e}^{u})\\&=\mathrm{e}^{u}\cos u+\mathrm{e}^{u}\sin u-\int\mathrm{e}^{u}\cos u\mathrm{d}u,\end{align*} $$

因此 $ \int e^{u}\cos u\mathrm{d}u=\frac{\mathrm{e}^{u}(\cos u+\sin u)}{2}+C $，故有

$$ \int\cos\ln x\mathrm{d}x=\frac{x(\cos\ln x+\sin\ln x)}{2}+C. $$

21.

$$ \begin{aligned}1.\ \int\left(\arcsin x\right)^{2}\mathrm{d}x&=x(\arcsin x)^{2}-\int\frac{2x\arcsin x}{\sqrt{1-x^{2}}}\mathrm{d}x\\&=x(\arcsin x)^{2}+\int2\arcsin x\mathrm{d}(\sqrt{1-x^{2}})\\&=x(\arcsin x)^{2}+2\sqrt{1-x^{2}}\arcsin x-2x+C.\\ \end{aligned} $$

22.

$$ \int\mathrm{e}^{x}\sin^{2}x\mathrm{d}x=\frac{1}{2}\int\mathrm{e}^{x}\left(1-\cos2x\right)\mathrm{d}x=\frac{1}{2}\mathrm{e}^{x}-\frac{1}{2}\int\mathrm{e}^{x}\cos2x\mathrm{d}x, $$

$$ \begin{align*}\int\mathrm{e}^{x}\cos2x\mathrm{d}x&=\int\cos2x\mathrm{d}(\mathrm{e}^{x})=\mathrm{e}^{x}\cos2x+2\int\mathrm{e}^{x}\sin2x\mathrm{d}x\\&=\mathrm{e}^{x}\cos2x+2\int\sin2x\mathrm{d}(\mathrm{e}^{x})\\&=\mathrm{e}^{x}\cos2x+2\mathrm{e}^{x}\sin2x-4\int\mathrm{e}^{x}\cos2x\mathrm{d}x,\end{align*} $$

得 $ \int e^{x}\cos2x dx=\frac{e^{x}\cos2x+2e^{x}\sin2x}{5}+C $，因此有

$$ \int\mathrm{e}^{x}\sin^{2}x\mathrm{d}x=\frac{1}{2}\mathrm{e}^{x}-\frac{1}{5}\mathrm{e}^{x}\sin2x-\frac{1}{10}\mathrm{e}^{x}\cos2x+C. $$

23.

$$ \begin{align*}\int x\ln^{2}x\mathrm{d}x&=\int\ln^{2}x\mathrm{d}\left(\frac{x^{2}}{2}\right)=\frac{x^{2}}{2}\ln^{2}x-\int x\ln x\mathrm{d}x\\&=\frac{x^{2}}{2}\ln^{2}x-\int\ln x\mathrm{d}\left(\frac{x^{2}}{2}\right)=\frac{x^{2}}{2}\ln^{2}x-\frac{x^{2}}{2}\ln x+\int\frac{x}{2}\mathrm{d}x\\&=\frac{x^{2}}{4}(2\ln^{2}x-2\ln x+1)+C.\end{align*} $$

1. 设 $ \sqrt{3x+9}=u $，即 $ x=\frac{1}{3}(u^{2}-9) $， $ \mathrm{d}x=\frac{2}{3}u\mathrm{d}u $，则

$$ \int\mathrm{e}^{\sqrt{3x+9}}\mathrm{d}x=\int\frac{2}{3}u\mathrm{e}^{u}\mathrm{d}u=\int\frac{2}{3}u\mathrm{d}(\mathrm{e}^{u}) $$

$$ \begin{aligned}&=\frac{2}{3}u\mathrm{e}^{u}-\int\frac{2}{3}\mathrm{e}^{u}\mathrm{d}u=\frac{2}{3}u\mathrm{e}^{u}-\frac{2}{3}\mathrm{e}^{u}+C\\&=\frac{2}{3}\mathrm{e}^{\sqrt{3x+9}}\left(\sqrt{3x+9}-1\right)+C.\end{aligned} $$

习题4-4

有理函数的积分

求下列不定积分：

1.

$$ \int\frac{x^{3}}{x+3}\mathrm{d}x. $$

2.

$$ \int\frac{2x+3}{x^{2}+3x-10}\mathrm{d}x. $$

3.

$$ \int\frac{x+1}{x^{2}-2x+5}\mathrm{d}x. $$

4.

$$ \int\frac{\mathrm{d}x}{x(x^{2}+1)}. $$

5.

$$ \int\frac{3}{x^{3}+1}\mathrm{d}x. $$

1. $ \int\frac{x^{2}+1}{\left(x+1\right)^{2}(x-1)} $dx.

1. $ \int\frac{x\mathrm{d}x}{(x+1)(x+2)(x+3)}. $

8.

$$ \int\frac{x^{5}+x^{4}-8}{x^{3}-x}\mathrm{d}x. $$

1. $ \int \frac{dx}{(x^{2} + 1)(x^{2} + x)} $.

10.

$$ \int\frac{1}{x^{4}-1}\mathrm{d}x. $$

11.

$$ \int\frac{\mathrm{d}x}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)}. $$

12.

$$ \int\frac{\left(x+1\right)^{2}}{\left(x^{2}+1\right)^{2}}\mathrm{d}x. $$

1. $ \int \frac{-x^{2}-2}{(x^{2}+x+1)^{2}}dx. $

14.

$$ \int\frac{\mathrm{d}x}{3+\sin^{2}x}. $$

15.

$$ \int\frac{\mathrm{d}x}{3+\cos x}. $$

16.

$$ \int\frac{\mathrm{d}x}{2+\sin x}. $$

17.

$$ \int\frac{\mathrm{d}x}{1+\sin x+\cos x}. $$

18.

$$ \int\frac{\mathrm{d}x}{2\sin x-\cos x+5}. $$

19.

$$ \int\frac{\mathrm{d}x}{1+\sqrt[3]{x+1}}. $$

20.

$$ \int\frac{\left(\sqrt{x}\right)^{3}-1}{\sqrt{x}+1}\mathrm{d}x. $$

21.

$$ \int\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\mathrm{d}x. $$

22.

$$ \int\frac{\mathrm{d}x}{\sqrt{x}+\sqrt[4]{x}}. $$

23.

$$ \int\sqrt{\frac{1-x}{1+x}}\frac{\mathrm{d}x}{x}. $$

24.

$$ \int\frac{\mathrm{d}x}{\sqrt[3]{\left(x+1\right)^{2}\left(x-1\right)^{4}}}. $$

解 1. $ \int \frac{x^3}{x + 3} \, dx = \int (x^2 - 3x + 9 - \frac{27}{x + 3}) \, dx $

$ = \frac{1}{3}x^3 - \frac{3}{2}x^2 + 9x - 27\ln|x + 3| + C. $

2.

$$ \int\frac{2x+3}{x^{2}+3x-10}\mathrm{d}x=\int\frac{\mathrm{d}(x^{2}+3x-10)}{x^{2}+3x-10}=\ln|x^{2}+3x-10|+C. $$

3.

$$ \begin{aligned}\int\frac{x+1}{x^{2}-2x+5}\mathrm{d}x&=\int\frac{x-1}{\left(x-1\right)^{2}+4}\mathrm{d}x+\frac{1}{2}\int\frac{1}{\left(\frac{x-1}{2}\right)^{2}+1}\mathrm{d}x\\&=\frac{1}{2}\ln\left(x^{2}-2x+5\right)+\arctan\frac{x-1}{2}+C.\end{aligned} $$

4.

$$ \begin{align*}\therefore\quad\int\frac{\mathrm{d}x}{x\left(x^{2}+1\right)}&=\int\left(\frac{1}{x}-\frac{x}{x^{2}+1}\right)\mathrm{d}x=\ln\left|x\right|-\frac{1}{2}\int\frac{\mathrm{d}\left(x^{2}+1\right)}{x^{2}+1}\\&=\ln\left|x\right|-\frac{1}{2}\ln\left(x^{2}+1\right)+C.\end{align*} $$

5.

$$ \begin{aligned}\int\frac{3}{1+x^{3}}\mathrm{d}x&=\int\frac{3}{\left(1+x\right)\left(x^{2}-x+1\right)}\mathrm{d}x=\int\left(\frac{1}{1+x}+\frac{2-x}{x^{2}-x+1}\right)\mathrm{d}x\\&=\ln|1+x|-\frac{1}{2}\int\frac{\left(x^{2}-x+1\right)^{\prime}}{x^{2}-x+1}\mathrm{d}x+\frac{3}{2}\int\frac{1}{x^{2}-x+1}\mathrm{d}x\\&=\ln|1+x|-\frac{1}{2}\ln(x^{2}-x+1)+\sqrt{3}\int\frac{1}{\left(\frac{2x-1}{\sqrt{3}}\right)^{2}+1}\mathrm{d}\left(\frac{2x-1}{\sqrt{3}}\right)\\&=\ln|1+x|-\frac{1}{2}\ln\left(x^{2}-x+1\right)+\sqrt{3}\arctan\frac{2x-1}{\sqrt{3}}+C.\end{aligned} $$

6.

$$ \begin{aligned}5.\ \int\frac{x^{2}+1}{\left(x+1\right)^{2}\left(x-1\right)}\mathrm{d}x&=\int\left[\frac{1}{2\left(x-1\right)}+\frac{1}{2\left(x+1\right)}-\frac{1}{\left(x+1\right)^{2}}\right]\mathrm{d}x\\&=\frac{1}{2}\ln\left|x-1\right|+\frac{1}{2}\ln\left|x+1\right|+\frac{1}{x+1}+C\\&=\frac{1}{2}\ln\left|x^{2}-1\right|+\frac{1}{x+1}+C.\end{aligned} $$

$$ \begin{aligned}7.\ \int\frac{x\mathrm{d}x}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}&=\int\left[\quad-\frac{1}{2\left(x+1\right)}+\frac{2}{x+2}-\frac{3}{2\left(x+3\right)}\right]\mathrm{d}x\\&=-\frac{1}{2}\ln\left|x+1\right|+2\ln\left|x+2\right|-\frac{3}{2}\ln\left|x+3\right|+C.\end{aligned} $$

$$ \begin{aligned}\cdot\int\frac{x^{5}+x^{4}-8}{x^{3}-x}\mathrm{d}x&=\int\left(x^{2}+x+1+\frac{8}{x}-\frac{3}{x-1}-\frac{4}{x+1}\right)\mathrm{d}x\\&=\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+8\ln|x|-3\ln|x-1|-4\ln|x+1|+C.\end{aligned} $$

$$ \begin{aligned}9.\ \int\frac{\mathrm{d}x}{\left(x^{2}+1\right)\left(x^{2}+x\right)}&=\int\left[\frac{1}{x}-\frac{1}{2\left(x+1\right)}-\frac{1+x}{2\left(x^{2}+1\right)}\right]\mathrm{d}x\\&=\ln|x|-\frac{1}{2}\ln|x+1|-\frac{1}{2}\arctan x-\frac{1}{4}\int\frac{\mathrm{d}(x^{2}+1)}{x^{2}+1}\\&=\ln|x|-\frac{1}{2}\ln|x+1|-\frac{1}{2}\arctan x-\frac{1}{4}\ln(x^{2}+1)+C.\\ \end{aligned} $$

10.

$$ \begin{aligned}\int\frac{1}{x^{4}-1}\mathrm{d}x&=\int\frac{1}{\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)}\mathrm{d}x\\&=\frac{1}{4}\int\frac{1}{x-1}\mathrm{d}x-\frac{1}{4}\int\frac{1}{x+1}\mathrm{d}x-\frac{1}{2}\int\frac{1}{x^{2}+1}\mathrm{d}x\\&=\frac{1}{4}\ln\left|\frac{x-1}{x+1}\right|-\frac{1}{2}\arctan x+C.\end{aligned} $$

11.

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)}&=\int\left(\frac{-x}{x^{2}+1}+\frac{x+1}{x^{2}+x+1}\right)\mathrm{d}x\\&=-\frac{\ln\left(x^{2}+1\right)}{2}+\frac{1}{2}\int\frac{\mathrm{d}\left(x^{2}+x+1\right)}{x^{2}+x+1}+\\&\quad\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}\mathrm{d}x\\&=-\frac{\ln\left(x^{2}+1\right)}{2}+\frac{\ln\left(x^{2}+x+1\right)}{2}+\frac{1}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}+C.\end{aligned} $$

12.

$$ \begin{aligned}\int\frac{\left(x+1\right)^{2}}{\left(x^{2}+1\right)^{2}}\mathrm{d}x&=\int\frac{x^{2}+1}{\left(x^{2}+1\right)^{2}}\mathrm{d}x+\int\frac{2x\mathrm{d}x}{\left(x^{2}+1\right)^{2}}\\&=\arctan x-\frac{1}{x^{2}+1}+C.\end{aligned} $$

13.

$$ \begin{aligned}&\int\frac{-x^{2}-2}{\left(x^{2}+x+1\right)^{2}}\mathrm{d}x=\int\left[\begin{array}{l}-\frac{1}{x^{2}+x+1}+\frac{x-1}{\left(x^{2}+x+1\right)^{2}}\end{array}\right]\mathrm{d}x\\&=-\int\frac{1}{x^{2}+x+1}\mathrm{d}x+\frac{1}{2}\int\frac{\mathrm{d}\left(x^{2}+x+1\right)}{\left(x^{2}+x+1\right)^{2}}-\frac{3}{2}\int\frac{1}{\left(x^{2}+x+1\right)^{2}}\mathrm{d}x,\end{aligned} $$

令 $ u = x + \frac{1}{2} $，并记 $ a = \frac{\sqrt{3}}{2} $，则

$$ \begin{aligned}\int\frac{1}{\left(x^{2}+x+1\right)^{2}}\mathrm{d}x&=\int\frac{1}{\left(u^{2}+a^{2}\right)^{2}}\mathrm{d}u=\frac{1}{2a^{2}}\bigg[\frac{u}{u^{2}+a^{2}}+\int\frac{1}{u^{2}+a^{2}}\mathrm{d}u\bigg]\\&=\frac{u}{2a^{2}\left(u^{2}+a^{2}\right)}+\frac{1}{2a^{2}}\int\frac{1}{u^{2}+a^{2}}\mathrm{d}u,\end{aligned} $$

由此得

$$ \begin{aligned}&\int\frac{1}{x^{2}+x+1}\mathrm{d}x+\frac{3}{2}\int\frac{1}{\left(x^{2}+x+1\right)^{2}}\mathrm{d}x\\=&\int\frac{1}{u^{2}+a^{2}}\mathrm{d}u+\frac{3}{2}\left[\frac{u}{2a^{2}\left(u^{2}+a^{2}\right)}+\frac{1}{2a^{2}}\int\frac{1}{u^{2}+a^{2}}\mathrm{d}u\right]\\=&\frac{3u}{4a^{2}\left(u^{2}+a^{2}\right)}+\left(\frac{3}{4a^{2}}+1\right)\int\frac{1}{u^{2}+a^{2}}\mathrm{d}u\\=&\frac{3u}{4a^{2}\left(u^{2}+a^{2}\right)}+\frac{1}{a}\bigg(\frac{3}{4a^{2}}+1\bigg)\arctan\frac{u}{a}+C_{1}\end{aligned} $$

$$ =\frac{2x+1}{2\left(x^{2}+x+1\right)}+\frac{4}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}+C_{1}, $$

因此有

$$ \begin{aligned}\int\frac{-x^{2}-2}{\left(x^{2}+x+1\right)^{2}}\mathrm{d}x&=-\frac{1}{2\left(x^{2}+x+1\right)}-\frac{2x+1}{2\left(x^{2}+x+1\right)}-\\&\quad\frac{4}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}+C\\&=-\frac{x+1}{x^{2}+x+1}-\frac{4}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}+C.\end{aligned} $$

14.

$$ \begin{align*}\int\frac{\mathrm{d}x}{3+\sin^{2}x}&=-\int\frac{\mathrm{d}(\cot x)}{3\csc^{2}x+1}\xlongequal{u=\cot x}-\int\frac{\mathrm{d}u}{3u^{2}+4}\\&=-\frac{1}{2\sqrt{3}}\arctan\frac{\sqrt{3}u}{2}+C\\&=-\frac{1}{2\sqrt{3}}\arctan\frac{\sqrt{3}\cot x}{2}+C.\end{align*} $$

1. 令 $ u = \tan \frac{x}{2} $，则

$$ \begin{align*}\int\frac{\mathrm{d}x}{3+\cos x}&=\int\frac{1}{3+\frac{1-u^{2}}{1+u^{2}}}\cdot\frac{2}{1+u^{2}}\mathrm{d}u=\int\frac{1}{2+u^{2}}\mathrm{d}u\\&=\frac{1}{\sqrt{2}}\arctan\frac{u}{\sqrt{2}}+C=\frac{1}{\sqrt{2}}\arctan\frac{\tan\frac{x}{2}}{\sqrt{2}}+C.\end{align*} $$

1. 令 $ u = \tan \frac{x}{2} $，则

$$ \begin{aligned}\int\frac{\mathrm{d}x}{2+\sin x}&=\int\frac{1}{2+\frac{2u}{1+u^{2}}}\cdot\frac{2}{1+u^{2}}\mathrm{d}u=\int\frac{1}{u^{2}+u+1}\mathrm{d}u&\\ &=\int\frac{1}{\left(u+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\mathrm{d}u=\frac{2}{\sqrt{3}}\arctan\frac{2u+1}{\sqrt{3}}+C\\ &\\ &=\frac{2}{\sqrt{3}}\arctan\frac{2\tan\frac{x}{2}+1}{\sqrt{3}}+C.\\ \end{aligned} $$

1. 令 $ u = \tan \frac{x}{2} $，则

$$ \int\frac{\mathrm{d}x}{1+\sin x+\cos x}=\int\frac{1}{1+\frac{2u}{1+u^{2}}+\frac{1-u^{2}}{1+u^{2}}}\cdot\frac{2}{1+u^{2}}\mathrm{d}u $$

$$ =\int\frac{\mathrm{d}u}{1+u}=\left.\ln\right|1+u\left|+\right.C=\left.\ln\right|1+\tan\frac{x}{2}\left|+\right.C. $$

1. 令 $ u = \tan \frac{x}{2} $，则

$$ \begin{align*}\int\frac{\mathrm{d}x}{2\sin x-\cos x+5}&=\int\frac{1}{\frac{4u}{1+u^{2}}-\frac{1-u^{2}}{1+u^{2}}+5}\cdot\frac{2}{1+u^{2}}\mathrm{d}u\\&=\int\frac{1}{3u^{2}+2u+2}\mathrm{d}u\\&=\frac{1}{3}\int\frac{1}{\left(u+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{5}}{3}\right)^{2}}\mathrm{d}\left(u+\frac{1}{3}\right)\\&=\frac{1}{\sqrt{5}}\arctan\frac{3u+1}{\sqrt{5}}+C\\&=\frac{1}{\sqrt{5}}\arctan\frac{3\tan\frac{x}{2}+1}{\sqrt{5}}+C.\end{align*} $$

1. 令 $ u = \sqrt[3]{x + 1} $，即 $ x = u^{3} - 1 $，则

$$ \begin{align*}\int\frac{\mathrm{d}x}{1+\sqrt[3]{x+1}}&=\int\frac{3u^{2}}{1+u}\mathrm{d}u=\int\left(3u-3+\frac{3}{1+u}\right)\mathrm{d}u\\&=\frac{3}{2}u^{2}-3u+3\ln|1+u|+C\\&=\frac{3}{2}\sqrt[3]{\left(x+1\right)^{2}}-3\sqrt[3]{x+1}+3\ln\left|1+\sqrt[3]{x+1}\right|+C.\end{align*} $$

20.

$$ \begin{aligned}0.\ \int\frac{\left(\sqrt{x}\right)^{3}-1}{\sqrt{x}+1}\mathrm{d}x&=\int\left(x-\sqrt{x}+1-\frac{2}{\sqrt{x}+1}\right)\mathrm{d}x\\&=\frac{x^{2}}{2}-\frac{2}{3}x\sqrt{x}+x-\int\frac{4t}{t+1}\mathrm{d}t\\&=\frac{x^{2}}{2}-\frac{2}{3}x\sqrt{x}+x-4\int\left(1-\frac{1}{t+1}\right)\mathrm{d}t\\&=\frac{x^{2}}{2}-\frac{2}{3}x\sqrt{x}+x-4\sqrt{x}+4\ln\left(\sqrt{x}+1\right)+C.\\ \end{aligned} $$

1. 令 $ u = \sqrt{x + 1} $，即 $ x = u^{2} - 1 $，则

$$ \begin{aligned}\int\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\mathrm{d}x&=\int\frac{u-1}{u+1}\cdot2u\mathrm{d}u=2\int\left(u-2+\frac{2}{u+1}\right)\mathrm{d}u\\&=u^{2}-4u+4\ln\left|u+1\right|+C\\&=x-4\sqrt{x+1}+4\ln\left(\sqrt{x+1}+1\right)+C_{1}\left(C_{1}=C+1\right).\end{aligned} $$

1. 令 $ u = \sqrt[4]{x} $，即 $ x = u^{4} $，则

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\sqrt{x}+\sqrt[4]{x}}&=\int\frac{1}{u^{2}+u}\cdot4u^{3}\mathrm{d}u=4\int\left(u-1+\frac{1}{u+1}\right)\mathrm{d}u\\&=2u^{2}-4u+4\ln|u+1|+C\\&=2\sqrt{x}-4\sqrt[4]{x}+4\ln(\sqrt[4]{x}+1)+C.\end{aligned} $$

1. 解法一

令 $ u=\sqrt{\frac{1-x}{1+x}} $，即 $ x=\frac{1-u^{2}}{1+u^{2}} $，则

$$ \begin{aligned}\int\sqrt{\frac{1-x}{1+x}}\cdot\frac{\mathrm{d}x}{x}&=\int u\cdot\frac{1+u^{2}}{1-u^{2}}\cdot\frac{-4u}{\left(1+u^{2}\right)^{2}}\mathrm{d}u=\int\frac{-4u^{2}}{\left(1-u^{2}\right)\left(1+u^{2}\right)}\mathrm{d}u\\&=\int\left(\frac{2}{1+u^{2}}-\frac{1}{1-u}-\frac{1}{1+u}\right)\mathrm{d}u\\&=2\arctan u+\ln|1-u|-\ln|1+u|+C\\&=2\arctan\sqrt{\frac{1-x}{1+x}}+\ln\left|\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right|+C.\end{aligned} $$

解法二

$$ \begin{aligned}\int\sqrt{\frac{1-x}{1+x}}\frac{\mathrm{d}x}{x}&=\int\frac{1-x}{x\sqrt{1-x^{2}}}\mathrm{d}x\xlongequal{x=\sin u}\int\frac{1-\sin u}{\sin u}\mathrm{d}u\\ &=\int\csc u\mathrm{d}u-\int\mathrm{d}u=\ln|\csc u-\cot u|-u+C\\ &=\ln\frac{1-\sqrt{1-x^{2}}}{|x|}-\arcsin x+C.\end{aligned} $$

24.

$$ \int\frac{\mathrm{d}x}{\sqrt[3]{\left(x+1\right)^{2}\left(x-1\right)^{4}}}=\int\frac{1}{x^{2}-1}\sqrt[3]{\frac{x+1}{x-1}}\mathrm{d}x, $$

令 $ u = \sqrt[3]{\frac{x + 1}{x - 1}} $，即 $ x = \frac{u^{3} + 1}{u^{3} - 1} $，得

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\sqrt[3]{\left(x+1\right)^{2}\left(x-1\right)^{4}}}&=\int\frac{u}{\left(\frac{u^{3}+1}{u^{3}-1}\right)^{2}-1}\cdot\frac{-6u^{2}}{\left(u^{3}-1\right)^{2}}\mathrm{d}u=-\frac{3}{2}\int\mathrm{d}u\\&=-\frac{3}{2}u+C=-\frac{3}{2\sqrt[3]{\frac{x+1}{x-1}}}+C.\end{aligned} $$

习题4-5

积分表的使用

利用积分表计算下列不定积分：

1.

$$ \int\frac{\mathrm{d}x}{\sqrt{4x^{2}-9}}. $$

2.

$$ \int\frac{1}{x^{2}+2x+5}\mathrm{d}x. $$

3.

$$ \int\frac{\mathrm{d}x}{\sqrt{5-4x+x^{2}}}. $$

4.

$$ \int\sqrt{2x^{2}+9}\mathrm{d}x. $$

5.

$$ \int\sqrt{3x^{2}-2}\mathrm{d}x. $$

6.

$$ \int\mathrm{e}^{2x}\cos x\mathrm{d}x. $$

7.

$$ \int x\arcsin\frac{x}{2}\mathrm{d}x. $$

8.

$$ \int\frac{\mathrm{d}x}{\left(x^{2}+9\right)^{2}}. $$

9.

$$ \int\frac{\mathrm{d}x}{\sin^{3}x}. $$

10.

$$ \int\mathrm{e}^{-2x}\sin3x\mathrm{d}x. $$

11.

$$ \int\sin3x\sin5x\mathrm{d}x. $$

12.

$$ \int\ln^{3}x\mathrm{d}x. $$

13.

$$ \int\frac{1}{x^{2}\left(1-x\right)}\mathrm{d}x. $$

14.

$$ \int\frac{\sqrt{x-1}}{x}\mathrm{d}x. $$

15.

$$ \int\frac{1}{\left(1+x^{2}\right)^{2}}\mathrm{d}x. $$

16.

$$ \int\frac{1}{x\sqrt{x^{2}-1}}\mathrm{d}x. $$

17.

$$ \int\frac{x}{\left(2+3x\right)^{2}}\mathrm{d}x. $$

18.

$$ \int\cos^{6}x\mathrm{d}x. $$

19.

$$ \int x^{2}\sqrt{x^{2}-2}\mathrm{d}x. $$

20.

$$ \int\frac{1}{2+5\cos x}\mathrm{d}x. $$

21.

$$ \int\frac{\mathrm{d}x}{x^{2}\sqrt{2x-1}}. $$

22.

$$ \int\sqrt{\frac{1-x}{1+x}}\mathrm{d}x. $$

23.

$$ \int\frac{x+5}{x^{2}-2x-1}\mathrm{d}x. $$

24.

$$ \int\frac{x\mathrm{d}x}{\sqrt{1+x-x^{2}}}. $$

25.

$$ \int\frac{x^{4}}{25+4x^{2}}\mathrm{d}x. $$

解 注意：下列各题中最后括号内所标的是所用积分公式在教材上册附录 IV 积分表中的编号.

1.

$$ \begin{align*}\int\frac{\mathrm{d}x}{\sqrt{4x^{2}-9}}&=\frac{1}{2}\int\frac{\mathrm{d}(2x)}{\sqrt{(2x)^{2}-3^{2}}}=\frac{1}{2}\ln\left|2x+\sqrt{(2x)^{2}-3^{2}}\right|+C\\&=\frac{1}{2}\ln\left|2x+\sqrt{4x^{2}-9}\right|+C.(45)\end{align*} $$

2.

$$ \mathrm{d}x=\int\frac{1}{x^{2}+2x+5}\mathrm{d}x=\int\frac{1}{\left(x+1\right)^{2}+2^{2}}\mathrm{d}(x+1)=\frac{1}{2}\arctan\frac{x+1}{2}+C. $$

3.

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\sqrt{5-4x+x^{2}}}&=\int\frac{\mathrm{d}(x-2)}{\sqrt{(x-2)^{2}+1}}=\ln[x-2+\sqrt{(x-2)^{2}+1}]+C\\&=\ln(x-2+\sqrt{5-4x+x^{2}})+C.(31)\end{aligned} $$

4.

$$ \begin{aligned}\int\sqrt{2x^{2}+9}\mathrm{d}x&=\frac{1}{\sqrt{2}}\int\sqrt{\left(\sqrt{2}x\right)^{2}+3^{2}}\mathrm{d}\left(\sqrt{2}x\right)\\&=\frac{1}{\sqrt{2}}\Biggl\{\frac{\sqrt{2}x}{2}\sqrt{\left(\sqrt{2}x\right)^{2}+3^{2}}+\frac{3^{2}}{2}\mathrm{ln}\bigl[\sqrt{2}x+\sqrt{\left(\sqrt{2}x\right)^{2}+3^{2}}\bigr]\Biggr\}+C\\&=\frac{x}{2}\sqrt{2x^{2}+9}+\frac{9\sqrt{2}}{4}\mathrm{ln}\bigl(\sqrt{2}x+\sqrt{2x^{2}+9}\bigr)\ +C.\left(39\right)\end{aligned} $$

5.

$$ \begin{aligned}\int\sqrt{3x^{2}-2}\mathrm{d}x&=\frac{1}{\sqrt{3}}\int\sqrt{\left(\sqrt{3}x\right)^{2}-\left(\sqrt{2}\right)^{2}}\mathrm{d}\left(\sqrt{3}x\right)\\&=\frac{1}{\sqrt{3}}\left[\frac{\sqrt{3}x}{2}\sqrt{\left(\sqrt{3}x\right)^{2}-\left(\sqrt{2}\right)^{2}}-\right.\\&\quad\left.\frac{\left(\sqrt{2}\right)^{2}}{2}\ln\left|\sqrt{3}x+\sqrt{\left(\sqrt{3}x\right)^{2}-\left(\sqrt{2}\right)^{2}}\right|\right]+C\\&=\frac{x}{2}\sqrt{3x^{2}-2}-\frac{\sqrt{3}}{3}\ln\left|\sqrt{3}x+\sqrt{3x^{2}-2}\right|+C.(\text{由}\ln x=0)\end{aligned} $$

6.

$$ \begin{align*}\int\mathrm{e}^{2x}\cos x\mathrm{d}x&=\frac{1}{2^{2}+1^{2}}\mathrm{e}^{2x}(\sin x+2\cos x)+C\\&=\frac{1}{5}\mathrm{e}^{2x}(\sin x+2\cos x)+C.(129)\end{align*} $$

7.

$$ \begin{align*}\int x\arcsin\frac{x}{2}\mathrm{d}x&=\left(\frac{x^{2}}{2}-\frac{2^{2}}{4}\right)\arcsin\frac{x}{2}+\frac{x}{4}\sqrt{2^{2}-x^{2}}+C\\&=\left(\frac{x^{2}}{2}-1\right)\arcsin\frac{x}{2}+\frac{x}{4}\sqrt{4-x^{2}}+C.\tag{114}\end{align*} $$

8.

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\left(x^{2}+9\right)^{2}}&=\int\frac{\mathrm{d}x}{\left(x^{2}+3^{2}\right)^{2}}\\&=\frac{x}{2\left(2-1\right)3^{2}\left(x^{2}+3^{2}\right)}+\frac{2\times2-3}{2\left(2-1\right)3^{2}}\int\frac{\mathrm{d}x}{x^{2}+3^{2}}\\&=\frac{x}{18\left(x^{2}+9\right)}+\frac{1}{18}\cdot\frac{1}{3}\arctan\frac{x}{3}+C\\&=\frac{x}{18\left(x^{2}+9\right)}+\frac{1}{54}\arctan\frac{x}{3}+C.\text{(20,19)}\end{aligned} $$

$$ \begin{aligned}9.\ \int\frac{\mathrm{d}x}{\sin^{3}x}&=-\frac{1}{2}\cdot\frac{\cos x}{\sin^{2}x}+\frac{1}{2}\int\frac{\mathrm{d}x}{\sin x}\\&=-\frac{\cos x}{2\sin^{2}x}+\frac{1}{2}\ln|\csc x-\cot x|+C.(97,88)\end{aligned} $$

10.

$$ \begin{aligned}0.\ \int\mathrm{e}^{-2x}\sin3x\mathrm{d}x&=\frac{1}{\left(-2\right)^{2}+3^{2}}\mathrm{e}^{-2x}\left(-2\sin3x-3\cos3x\right)+C\\&=-\frac{\mathrm{e}^{-2x}}{13}(2\sin3x+3\cos3x)+C.(128)\end{aligned} $$

11.

$$ \begin{aligned}\int\sin3x\sin5x\mathrm{d}x&=-\frac{1}{2(3+5)}\sin(3+5)x+\frac{1}{2(3-5)}\sin(3-5)x+C\\&=-\frac{1}{16}\sin8x+\frac{1}{4}\sin2x+C.(101)\end{aligned} $$

12.

$$ \begin{align*}\int\ln^{3}x\mathrm{d}x&=x(\ln x)^{3}-3\int\ln^{2}x\mathrm{d}x\\&=x(\ln x)^{3}-3\Big[x(\ln x)^{2}-2\int\ln x\mathrm{d}x\Big]\\&=x(\ln x)^{3}-3x(\ln x)^{2}+6\int\ln x\mathrm{d}x\\&=x(\ln x)^{3}-3x(\ln x)^{2}+6(x\ln x-x)+C\\&=x\ln^{3}x-3x\ln^{2}x+6x\ln x-6x+C.(135,132)\end{align*} $$

13.

$$ \int\frac{1}{x^{2}\left(1-x\right)}\mathrm{d}x=-\left.\frac{1}{x}\right.-\ln\left|\frac{1-x}{x}\right|+C.\left(6\right) $$

14.

$$ \begin{aligned}\cdot\int\frac{\sqrt{x-1}}{x}\mathrm{d}x&=2\sqrt{x-1}-\int\frac{1}{x\sqrt{x-1}}\mathrm{d}x\\&=2\sqrt{x-1}-2\arctan\sqrt{x-1}+C.(17,15)\end{aligned} $$

15.

$$ \begin{aligned}\int\frac{1}{\left(1+x^{2}\right)^{2}}\mathrm{d}x&=\frac{x}{2\left(1+x^{2}\right)}+\frac{1}{2}\int\frac{1}{1+x^{2}}\mathrm{d}x\\&=\frac{x}{2\left(1+x^{2}\right)}+\frac{1}{2}\arctan x+C.(20,19)\end{aligned} $$

16.

$$ \int\frac{1}{x\sqrt{x^{2}-1}}\mathrm{d}x=\arccos\frac{1}{\mid x\mid}+C.(51) $$

17.

$$ \mathrm{~\boldmath~\cdot~}\int\frac{x}{\left(2+3x\right)^{2}}\mathrm{d}x\;=\;\frac{1}{9}\Big(\ln\mid2+3x\mid+\frac{2}{2+3x}\Big)+C.\left(7\right) $$

18.

$$ \begin{aligned}\int\cos^{6}x\mathrm{d}x&=\frac{1}{6}\cos^{5}x\sin x+\frac{5}{6}\int\cos^{4}x\mathrm{d}x\\&=\frac{1}{6}\cos^{5}x\sin x+\frac{5}{6}\bigg(\frac{1}{4}\cos^{3}x\sin x+\frac{3}{4}\int\cos^{2}x\mathrm{d}x\bigg)\\&=\frac{1}{6}\cos^{5}x\sin x+\frac{5}{24}\cos^{3}x\sin x+\frac{5}{8}\int\cos^{2}x\mathrm{d}x\\&=\frac{1}{6}\cos^{5}x\sin x+\frac{5}{24}\cos^{3}x\sin x+\frac{5}{8}\bigg(\frac{1}{2}\cos x\sin x+\frac{1}{2}\int\mathrm{d}x\bigg)\\&=\frac{1}{6}\cos^{5}x\sin x+\frac{5}{24}\cos^{3}x\sin x+\frac{5}{16}\cos x\sin x+\frac{5}{16}x+C.(96)\end{aligned} $$

19.

$$ \begin{aligned}9.\int x^{2}\sqrt{x^{2}-2}\mathrm{d}x&=\frac{x}{8}(2x^{2}-2)\sqrt{x^{2}-2}-\frac{4}{8}\ln\left|x+\sqrt{x^{2}-2}\right|+C\\&=\frac{x}{4}(x^{2}-1)\sqrt{x^{2}-2}-\frac{1}{2}\ln\left|x+\sqrt{x^{2}-2}\right|+C.(56)\end{aligned} $$

20.

$$ \begin{aligned}\int\frac{1}{2+5\cos x}\mathrm{d}x&=\frac{1}{7}\sqrt{\frac{7}{3}}\ln\left|\frac{\tan\frac{x}{2}+\sqrt{\frac{7}{3}}}{\tan\frac{x}{2}-\sqrt{\frac{7}{3}}}\right|+C\\&=\frac{1}{\sqrt{21}}\ln\left|\frac{\sqrt{3}\tan\frac{x}{2}+\sqrt{7}}{\sqrt{3}\tan\frac{x}{2}-\sqrt{7}}\right|+C.(106)\end{aligned} $$

21.

$$ \begin{aligned}\int\frac{\mathrm{d}x}{x^{2}\sqrt{2x-1}}&=-\left.\frac{\sqrt{2x-1}}{-x}-\frac{2}{-2}\right]\int\frac{\mathrm{d}x}{x\sqrt{2x-1}}\\&=\frac{\sqrt{2x-1}}{x}+2\arctan\sqrt{2x-1}+C.(16,15)\end{aligned} $$

1. 解法一

$$ \begin{aligned}\int\sqrt{\frac{1-x}{1+x}}\mathrm{d}x&=\int\frac{1-x}{\sqrt{1-x^{2}}}\mathrm{d}x=\int\frac{1}{\sqrt{1-x^{2}}}\mathrm{d}x-\int\frac{x}{\sqrt{1-x^{2}}}\mathrm{d}x\\&=\arcsin x+\sqrt{1-x^{2}}+C.(59,61)\end{aligned} $$

解法二

$$ \begin{aligned}\int_{\sqrt{\frac{1-x}{1+x}}}\mathrm{d}x&=\left.(x+1)\sqrt{\frac{1-x}{1+x}}-2\arcsin\sqrt{\frac{1-x}{2}}+C\right.\\&=\sqrt{1-x^{2}}-2\arcsin\sqrt{\frac{1-x}{2}}+C.(80)\end{aligned} $$

23.

$$ \begin{aligned}3.\ \int\frac{x+5}{x^{2}-2x-1}\mathrm{d}x&=\int\frac{x}{x^{2}-2x-1}\mathrm{d}x+5\int\frac{1}{x^{2}-2x-1}\mathrm{d}x\\&=\frac{1}{2}\ln|x^{2}-2x-1|-\frac{-2}{2}\int\frac{1}{x^{2}-2x-1}\mathrm{d}x+5\int\frac{1}{x^{2}-2x-1}\mathrm{d}x\\&=\frac{1}{2}\ln|x^{2}-2x-1|+6\cdot\frac{1}{\sqrt{\left(-2\right)^{2}-4\cdot1\cdot\left(-1\right)}}\\&\quad\ln\left|\frac{2x-2-\sqrt{\left(-2\right)^{2}-4\cdot1\cdot\left(-1\right)}}{2x-2+\sqrt{\left(-2\right)^{2}-4\cdot1\cdot\left(-1\right)}}\right|+C\\&=\frac{1}{2}\ln|x^{2}-2x-1|+\frac{3}{\sqrt{2}}\ln\left|\frac{x-\left(\sqrt{2}+1\right)}{x+\left(\sqrt{2}-1\right)}\right|+C.\ (30,29)\end{aligned} $$

24.

$$ \int\frac{x\mathrm{d}x}{\sqrt{1+x-x^{2}}}=-\sqrt{1+x-x^{2}}+\frac{1}{2}\arcsin\frac{2x-1}{\sqrt{5}}+C.(78) $$

25.

$$ \begin{aligned}\int\frac{x^{4}}{25+4x^{2}}\mathrm{d}x&=\int\bigg(\frac{1}{4}x^{2}-\frac{25}{16}+\frac{625}{16}\cdot\frac{1}{25+4x^{2}}\bigg)\mathrm{d}x\\&=\frac{x^{3}}{12}-\frac{25}{16}x+\frac{625}{32}\int\frac{1}{5^{2}+(2x)^{2}}\mathrm{d}(2x)\end{aligned} $$

$$ \begin{aligned}&=\frac{x^{3}}{12}-\frac{25}{16}x+\frac{625}{32}\cdot\frac{1}{5}\arctan\frac{2x}{5}+C\\&=\frac{x^{3}}{12}-\frac{25}{16}x+\frac{125}{32}\arctan\frac{2x}{5}+C.(19)\end{aligned} $$

总习题四

1. 填空：

(1) $ \int x^{3} e^{x} dx = $ ___;

(2) $ \int \frac{x + 5}{x^2 - 6x + 13} \, dx = $ ___.

解（1）

$$ \begin{align*}\int x^{3}\mathrm{e}^{x}\mathrm{d}x&=\int x^{3}\mathrm{d}(\mathrm{e}^{x})=x^{3}\mathrm{e}^{x}-3\int x^{2}\mathrm{d}(\mathrm{e}^{x})\\&=x^{3}\mathrm{e}^{x}-3\Big[x^{2}\mathrm{e}^{x}-\int2x\mathrm{d}(\mathrm{e}^{x})\Big]=x^{3}\mathrm{e}^{x}-3x^{2}\mathrm{e}^{x}+6\Big(x\mathrm{e}^{x}-\int\mathrm{e}^{x}\mathrm{d}x\Big)\\&=x^{3}\mathrm{e}^{x}-3x^{2}\mathrm{e}^{x}+6x\mathrm{e}^{x}-6\mathrm{e}^{x}+C,\end{align*} $$

因此，应填 $ x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+C. $

(2)

$$ \begin{aligned}\int\frac{x+5}{x^{2}-6x+13}\mathrm{d}x&=\frac{1}{2}\int\frac{\left(x^{2}-6x+13\right)^{\prime}}{x^{2}-6x+13}\mathrm{d}x+\int\frac{8}{x^{2}-6x+13}\mathrm{d}x\\&=\frac{1}{2}\ln(x^{2}-6x+13)+\int\frac{8}{\left(x-3\right)^{2}+4}\mathrm{d}x\\&=\frac{1}{2}\ln(x^{2}-6x+13)+4\arctan\frac{x-3}{2}+C,\end{aligned} $$

因此，应填 $ \frac{1}{2}\ln\left(x^{2}-6x+13\right)+4\arctan\frac{x-3}{2}+C $

1. 以下两题中给出了四个结论，从中选出一个正确的结论：

(1) 已知 $ f'(x) = \frac{1}{x(1 + 2\ln x)} $，且 $ f(1) = 1 $，则 $ f(x) $ 等于（）.

(A) $ \ln(1 + 2\ln x) + 1 $

(B) $ \frac{1}{2}\ln(1 + 2\ln x) + 1 $

(C) $ \frac{1}{2}\ln(1 + 2\ln x) + \frac{1}{2} $

(D) $ 2\ln(1 + 2\ln x) + 1 $

(2) 在下列等式中, 正确的结果是().

(A) $ \int f'(x) \, \mathrm{d}x = f(x) $ (B) $ \int \mathrm{d}f(x) = f(x) $

(C) $ \frac{d}{dx}\int f(x) \, dx = f(x) $ (D) $ \int f(x) \, dx = f(x) $

解 （1）由微积分基本定理，有

$$ \begin{align*}f(x)-f(1)&=\int_{1}^{x}f^{\prime}(t)\mathrm{d}t=\int_{1}^{x}\frac{1}{t(1+2\ln t)}\mathrm{d}t=\frac{1}{2}\int_{1}^{x}\frac{1}{1+2\ln t}\mathrm{d}(1+2\ln t)\\&=\frac{1}{2}\big[\ln(1+2\ln t)\big]_{1}^{x}=\frac{1}{2}\ln(1+2\ln x),\end{align*} $$

根据条件 $ f(1) = 1 $ , 得 $ f(x) = \frac{1}{2}\ln(1 + 2\ln x) + 1 $ 。故选 (B).

（2）根据微分运算与积分运算的关系，可知

$$ \int\mathrm{d}\boldsymbol{f}(\boldsymbol{x})=\int\boldsymbol{f}^{\prime}(\boldsymbol{x})\mathrm{d}\boldsymbol{x}=\boldsymbol{f}(\boldsymbol{x})+\boldsymbol{C}, $$

$$ \frac{\mathrm{d}}{\mathrm{d}x}\int f(x)\mathrm{d}x=f(x), $$

$$ \mathrm{d}\int f(x)\mathrm{d}x=\left(\frac{\mathrm{d}}{\mathrm{d}x}\int f(x)\mathrm{d}x\right)\mathrm{d}x=f(x)\mathrm{d}x, $$

故选(C).

1. 已知 $ \frac{\sin x}{x} $是 $ f(x) $的一个原函数，求 $ \int x^{3}f'(x)dx $.

解 根据条件，有 $ \int f(x)dx = \frac{\sin x}{x} + C $，即 $ f(x) = \left(\frac{\sin x}{x}\right)^{\prime} = \frac{x\cos x - \sin x}{x^{2}} $，因

此

$$ \begin{align*}\int x^{3}f^{\prime}(x)\mathrm{d}x&=x^{3}f(x)-\int3x^{2}f(x)\mathrm{d}x=x(x\cos x-\sin x)-3\int x^{2}\mathrm{d}\biggl(\frac{\sin x}{x}\biggr)\\&=x^{2}\cos x-x\sin x-3\biggl(x^{2}\cdot\frac{\sin x}{x}-\int\frac{\sin x}{x}\cdot2x\mathrm{d}x\biggr)\\&=x^{2}\cos x-4x\sin x-6\cos x+C.\end{align*} $$

1. 求下列不定积分（其中 a, b 为常数）：

(1) $ \int \frac{dx}{e^x - e^{-x}}; $

(2) $ \int\frac{x}{(1-x)^{3}}\mathrm{d}x $;

(3) $ \int \frac{x^{2}}{a^{6} - x^{6}} \mathrm{d}x (a > 0) $

$$ \int\frac{1+\cos x}{x+\sin x}\mathrm{d}x $$

(5) $ \int \frac{\ln\ln x}{x} dx $;

(6) $ \int \frac{\sin x \cos x}{1 + \sin^{4} x} \, dx $;

(7) $ \int \tan^{4} x \, dx $;

(8) $ \int \sin x \sin 2x \sin 3x dx $;

(9) $ \int \frac{dx}{x(x^{6} + 4)}; $

(10) $ \int\sqrt{\frac{a + x}{a - x}}dx $ (a > 0);

(11) $ \int \frac{dx}{\sqrt{x(1+x)}} $;

(12) $ \int x \cos^{2} x \, dx $;

(13) $ \int e^{ax} \cos bx \, dx $;

$$ \int\frac{\mathrm{d}x}{\sqrt{1+\mathrm{e}^{x}}}; $$

(15) $ \int\frac{\mathrm{d}x}{x^{2}\sqrt{x^{2}-1}}; $

(16) $ \int \frac{dx}{(a^2 - x^2)^{5/2}} $;

(17) $ \int \frac{dx}{x^4 \sqrt{1 + x^2}}; $

(18) $ \int \sqrt{x} \sin \sqrt{x} \, dx $;

(19) $ \int \ln(1 + x^{2}) \, dx $;

(20) $ \int\frac{\sin^{2}x}{\cos^{3}x}dx $;

(21) $ \int \arctan \sqrt{x} \, dx $;

(22) $ \int \frac{\sqrt{1 + \cos x}}{\sin x} \, dx $;

(23) $ \int \frac{x^{3}}{(1 + x^{8})^{2}} \mathrm{d}x $;

(24) $ \int \frac{x^{11}}{x^{8} + 3x^{4} + 2} \, dx $;

(25) $ \int\frac{dx}{16 - x^{4}} $;

(26) $ \int \frac{\sin x}{1 + \sin x} \, dx $;

(27) $ \int \frac{x + \sin x}{1 + \cos x} \, dx $;

(28) $ \int e^{\sin x} \frac{x \cos^{3} x - \sin x}{\cos^{2} x} dx $;

(29) $ \int \frac{\sqrt[3]{x}}{x(\sqrt{x} + \sqrt[3]{x})} \, dx $;

(30) $ \int \frac{dx}{(1 + e^{x})^2}; $

(31) $ \int \frac{e^{3x} + e^{x}}{e^{4x} - e^{2x} + 1} \, dx $;

(32) $ \int \frac{x e^{x}}{(e^{x} + 1)^{2}} \, dx $;

(33) $ \int \ln^{2}(x + \sqrt{1 + x^{2}}) \, dx $;

(34) $ \int \frac{\ln x}{(1 + x^{2})^{\frac{3}{2}}} dx $;

(35) $ \int \sqrt{1 - x^2} \arcsin x \, dx $;

$ \int \frac{x^3 \arccos x}{\sqrt{1 - x^2}} \, dx $;

(37) $ \int \frac{\cot x}{1 + \sin x} \, dx $;

(38) $ \int \frac{dx}{\sin^3 x \cos x} $;

(39) $ \int \frac{dx}{(2 + \cos x) \sin x}; $

$$ \int\frac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x. $$

解 (1) $ \int \frac{dx}{e^x - e^{-x}} = \int \frac{e^x dx}{e^{2x} - 1} = \frac{1}{2} \int \left( \frac{1}{e^x - 1} - \frac{1}{e^x + 1} \right) d(e^x) $

$ = \frac{1}{2} \ln \frac{|\mathbf{e}^x - \mathbf{1}|}{\mathbf{e}^x + 1} + C. $

(2) $ \int \frac{x}{(1 - x)^3} \, \mathrm{d}x \xlongequal{u = 1 - x} \int \left( \frac{1}{u^2} - \frac{1}{u^3} \right) \mathrm{d}u = -\frac{1}{u} + \frac{1}{2u^2} + C $

$ = -\frac{1}{1 - x} + \frac{1}{2(1 - x)^2} + C. $

(3) $ \int \frac{x^2}{a^6 - x^6} dx = \int \frac{d(x^3)}{3(a^6 - x^6)} \xrightarrow{u = x^3} \int \frac{du}{3(a^6 - u^2)} $

$ = \frac{1}{6a^3} \int \left( \frac{1}{a^3 + u} + \frac{1}{a^3 - u} \right) du $

$$ \begin{aligned}=\ \frac{1}{6a^{3}}\ln\left|\frac{a^{3}\ +u}{a^{3}\ -u}\right|+C\ =\ \frac{1}{6a^{3}}\ln\left|\frac{a^{3}\ +x^{3}}{a^{3}\ -x^{3}}\right|+C.\end{aligned} $$

(4)

$$ \int\frac{1+\cos x}{x+\sin x}\mathrm{d}x=\int\frac{\mathrm{d}(x+\sin x)}{x+\sin x}=\ln1x+\sin x\mid+C. $$

(5)

$$ \begin{aligned}\int\frac{\ln\ln x}{x}\mathrm{d}x&=\int\ln\ln x\mathrm{d}(\ln x)=\ln x\ln\ln x-\int\ln x\cdot\frac{1}{x\ln x}\mathrm{d}x\\&=\ln x(\ln\ln x-1)+C.\end{aligned} $$

(6)

$$ \int\frac{\sin x\cos x}{1+\sin^{4}x}\mathrm{d}x=\frac{1}{2}\int\frac{\mathrm{d}(\sin^{2}x)}{1+\sin^{4}x}=\frac{\arctan(\sin^{2}x)}{2}+C. $$

(7)

$$ \begin{align*}\int\tan^{4}x\mathrm{d}x&=\int\tan^{2}x\left(\sec^{2}x-1\right)\mathrm{d}x\\&=\int\tan^{2}x\mathrm{d}(\tan x)-\int(\sec^{2}x-1)\mathrm{d}x\\&=\frac{1}{3}\tan^{3}x-\tan x+x+C.\end{align*} $$

(8)

$$ \begin{align*}\int\sin x\sin2x\sin3x\mathrm{d}x&=\int\frac{1}{2}(\cos x-\cos3x)\sin3x\mathrm{d}x\\&=\frac{1}{2}\int\cos x\sin3x\mathrm{d}x-\frac{1}{2}\int\cos3x\sin3x\mathrm{d}x\\&=\frac{1}{4}\int(\sin2x+\sin4x)\mathrm{d}x-\frac{1}{12}\sin^{2}3x\\&=-\frac{1}{16}\cos4x-\frac{1}{8}\cos2x-\frac{1}{12}\sin^{2}3x+C.\end{align*} $$

（9）

$$ \begin{aligned}\int\frac{\mathrm{d}x}{x\left(x^{6}+4\right)}&\xlongequal{x\ =\frac{1}{u}}\int\frac{-u^{5}\mathrm{d}u}{1+4u^{6}}=-\frac{1}{24}\int\frac{\mathrm{d}(1+4u^{6})}{1+4u^{6}}\\&=-\frac{1}{24}\ln(1+4u^{6})+C=-\frac{1}{24}\ln\frac{x^{6}+4}{x^{6}}+C\\&=\frac{1}{4}\ln|x|-\frac{1}{24}\ln(x^{6}+4)+C.\end{aligned} $$

（10）解法一

$$ \begin{align*}\int\sqrt{\frac{a+x}{a-x}}\mathrm{d}x&=\int\frac{a+x}{\sqrt{a^{2}-x^{2}}}\mathrm{d}x=a\int\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}}\mathrm{d}\left(\frac{x}{a}\right)-\frac{1}{2}\int\frac{\mathrm{d}(a^{2}-x^{2})}{\sqrt{a^{2}-x^{2}}}\\&=a\arcsin\frac{x}{a}-\sqrt{a^{2}-x^{2}}+C.\end{align*} $$

解法二 令 $ u = \sqrt{\frac{a + x}{a - x}} $，即 $ x = a \frac{u^{2} - 1}{u^{2} + 1} $，则

$$ \int_{\sqrt{\frac{a+x}{a-x}}}\mathrm{d}x=\int u\cdot\frac{4au}{\left(1+u^{2}\right)^{2}}\mathrm{d}u=\int-2a u\mathrm{d}\left(\frac{1}{1+u^{2}}\right) $$

$$ \begin{aligned}&=-\frac{2au}{1+u^{2}}+\int\frac{2a}{1+u^{2}}\mathrm{d}u\\&=-\frac{2au}{1+u^{2}}+2a\arctan u+C\\&=\left(x-a\right)\sqrt{\frac{a+x}{a-x}}+2a\arctan\sqrt{\frac{a+x}{a-x}}+C\\&=-\sqrt{a^{2}-x^{2}}+2a\arctan\sqrt{\frac{a+x}{a-x}}+C.\end{aligned} $$

（11）解法一

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\sqrt{x(1+x)}}&=\int\frac{\mathrm{d}x}{\sqrt{\left(x+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}\\&\xlongequal{x=-\frac{1}{2}+\frac{1}{2}\sec u}\int\sec u\mathrm{d}u=\ln|\sec u+\tan u|+C\\&=\ln\left|2x+1+2\sqrt{x(1+x)}\right|+C.\end{aligned} $$

解法二 当 x > 0 时，因为 $ \frac{1}{\sqrt{x(1+x)}} = \frac{1}{x} \sqrt{\frac{x}{1+x}} $，故令 $ u = \sqrt{\frac{x}{1+x}} $，即 $ x = \frac{u^{2}}{1 - u^{2}} $，则

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\sqrt{x(1+x)}}&=\int\frac{2}{1-u^{2}}\mathrm{d}u=\int\biggl(\frac{1}{1-u}+\frac{1}{1+u}\biggr)\mathrm{d}u\\&=\ln\left|\frac{1+u}{1-u}\right|+C=\ln\left|\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}\right|+C\\&=\ln\left|2x+1+2\sqrt{x(1+x)}\right|+C,\end{aligned} $$

当 x < -1 时，同样可得 $ \int \frac{dx}{\sqrt{x(1+x)}} = \ln |2x + 1 + 2\sqrt{x(1+x)}| + C. $

(12)

$$ \begin{align*}\int x\cos^{2}x\mathrm{d}x&=\frac{1}{2}\int x(1+\cos2x)\mathrm{d}x=\frac{1}{4}\int x\mathrm{d}(2x+\sin2x)\\&=\frac{x(2x+\sin2x)}{4}-\frac{1}{4}\int(2x+\sin2x)\mathrm{d}x\\&=\frac{x^{2}}{4}+\frac{x\sin2x}{4}+\frac{\cos2x}{8}+C.\end{align*} $$

（13）当 $ a\neq0 $时，

$$ \begin{aligned}\int\mathrm{e}^{ax}\cos bx\mathrm{d}x&=\frac{1}{a}\int\cos bx\mathrm{d}(\mathrm{e}^{ax})\\&=\frac{1}{a}\mathrm{e}^{ax}\cos bx+\frac{b}{a}\int\mathrm{e}^{ax}\sin bx\mathrm{d}x\end{aligned} $$

$$ \begin{aligned}&=\frac{1}{a}\mathrm{e}^{ax}\cos bx+\frac{b}{a^{2}}\int\sin bxd(\mathrm{e}^{ax})\\&=\frac{1}{a}\mathrm{e}^{ax}\cos bx+\frac{b}{a^{2}}\mathrm{e}^{ax}\sin bx-\frac{b^{2}}{a^{2}}\int\mathrm{e}^{ax}\cos bxdx.\end{aligned} $$

因此有

$$ \int\mathrm{e}^{ax}\cos\ b x\mathrm{d}x\ =\frac{1}{a^{2}\ +\ b^{2}}\mathrm{e}^{ax}\left(\ a\cos\ b x\ +\ b\sin\ b x\right)\ +\ C, $$

当a=0时，

$$ \int\mathrm{e}^{ax}\cos bx\mathrm{d}x=\{\begin{aligned}\frac{\sin bx}{b}+C,\quad&b\neq0,\\ x+C,\quad&b=0.\end{aligned}. $$

（14）令 $ u = \sqrt{1 + e^{x}} $，即作换元 $ x = \ln(u^{2} - 1) $，得

$$ \int\frac{\mathrm{d}x}{\sqrt{1+\mathrm{e}^{x}}}=\int\frac{2\mathrm{d}u}{u^{2}-1}=\ln\left|\frac{u-1}{u+1}\right|+C=\ln\frac{\sqrt{1+\mathrm{e}^{x}}-1}{\sqrt{1+\mathrm{e}^{x}}+1}+C. $$

（15）

$$ \int\frac{\mathrm{d}x}{x^{2}\sqrt{x^{2}-1}}\xlongequal{x=\frac{1}{u}}-\int\frac{u\mathrm{d}u}{\sqrt{1-u^{2}}}=\sqrt{1-u^{2}}+C=\frac{\sqrt{x^{2}-1}}{x}+C, $$

易知当x<0和x>0时的结果相同.

（16）设 $ x = a \sin u \left( -\frac{\pi}{2} < u < \frac{\pi}{2} \right) $，则 $ \sqrt{a^{2} - x^{2}} = a \cos u, \mathrm{d}x = a \cos u \mathrm{d}u $，于是

$$ \begin{align*}\int\frac{\mathrm{d}x}{\left(a^{2}-x^{2}\right)^{5/2}}&=\frac{1}{a^{4}}\int\sec^{4}u\mathrm{d}u=\frac{1}{a^{4}}\int\left(\tan^{2}u+1\right)\mathrm{d}(\tan u)\\&=\frac{\tan^{3}u}{3a^{4}}+\frac{\tan u}{a^{4}}+C\\&=\frac{1}{3a^{4}}\bigg[\frac{x^{3}}{\sqrt{(a^{2}-x^{2})^{3}}}+\frac{3x}{\sqrt{a^{2}-x^{2}}}\bigg]+C.\end{align*} $$

（17）

$$ \begin{aligned}\int\frac{\mathrm{d}x}{x^{4}\sqrt{1+x^{2}}}&\xlongequal{x=\frac{1}{u}}\int\frac{-u^{3}\mathrm{d}u}{\sqrt{1+u^{2}}}=-\int\left(u\sqrt{1+u^{2}}-\frac{u}{\sqrt{1+u^{2}}}\right)\mathrm{d}u\\&=-\frac{1}{3}\left(1+u^{2}\right)^{\frac{3}{2}}+\sqrt{1+u^{2}}+C\\&=-\frac{1}{3}\frac{\sqrt{\left(1+x^{2}\right)^{3}}}{x^{3}}+\frac{\sqrt{1+x^{2}}}{x}+C,\end{aligned} $$

易知当x<0和x>0时结果相同.

(18)

$$ \begin{aligned}\int\sqrt{x}\sin\sqrt{x}\mathrm{d}x&\xlongequal{x=u^{2}}\int2u^{2}\sin u\mathrm{d}u=-\int2u^{2}\mathrm{d}(\cos u)\\&=-2u^{2}\cos u+\int4u\cos u\mathrm{d}u\\&=-2u^{2}\cos u+\int4u\mathrm{d}(\sin u)\\&=-2u^{2}\cos u+4u\sin u-\int4\sin u\mathrm{d}u\\&=-2u^{2}\cos u+4u\sin u+4\cos u+C\\&=-2x\cos\sqrt{x}+4\sqrt{x}\sin\sqrt{x}+4\cos\sqrt{x}+C.\end{aligned} $$

（19）

$$ \begin{aligned}\int\ln\left(1+x^{2}\right)\mathrm{d}x&=\left.x\ln\left(1+x^{2}\right)\right.-\int\frac{2x^{2}}{1+x^{2}}\mathrm{d}x\\&=\left.x\ln\left(1+x^{2}\right)\right.-2x+2\arctan x+C.\end{aligned} $$

(20)

$$ \begin{align*}\int\frac{\sin^{2}x}{\cos^{3}x}\mathrm{d}x&=\int\tan^{2}x\sec x\mathrm{d}x=\int\sec^{3}x\mathrm{d}x-\int\sec x\mathrm{d}x\\&=\left(\frac{1}{2}\sec x\tan x+\frac{1}{2}\int\sec x\mathrm{d}x\right)-\int\sec x\mathrm{d}x\\&=\frac{1}{2}\sec x\tan x-\frac{1}{2}\int\sec x\mathrm{d}x\\&=\frac{1}{2}\sec x\tan x-\frac{1}{2}\ln|\sec x+\tan x|+C.\end{align*} $$

(21)

$$ \begin{align*}\int\arctan\sqrt{x}\mathrm{d}x&=\int\arctan\sqrt{x}\mathrm{d}(1+x)=(1+x)\arctan\sqrt{x}-\int\frac{1}{2\sqrt{x}}\mathrm{d}x\\ &=(1+x)\arctan\sqrt{x}-\sqrt{x}+C.\end{align*} $$

(22)

$$ \begin{align*}\int\frac{\sqrt{1+\cos x}}{\sin x}\mathrm{d}x&=\int\frac{\sqrt{2}\left|\cos\frac{x}{2}\right|}{2\sin\frac{x}{2}\cos\frac{x}{2}}\mathrm{d}x=\pm\sqrt{2}\int\csc\frac{x}{2}\mathrm{d}\left(\frac{x}{2}\right)\\&=\pm\sqrt{2}\ln\left|\csc\frac{x}{2}-\cot\frac{x}{2}\right|+C,\end{align*} $$

上式当 $ \cos\frac{x}{2}>0 $时取正，当 $ \cos\frac{x}{2}<0 $时取负.

$$ \begin{aligned} 当 \cos\frac{x}{2}>0 时 ,\ln\left|\csc\frac{x}{2}-\cot\frac{x}{2}\right|&=\ln\frac{1-\cos\frac{x}{2}}{\left|\sin\frac{x}{2}\right|}\\&=\ln\left(\left|\csc\frac{x}{2}\right|-\left|\cot\frac{x}{2}\right|\right),\end{aligned} $$

$$ \begin{aligned} 当 \cos\frac{x}{2}&<0 时 ,\ln\left|\csc\frac{x}{2}-\cot\frac{x}{2}\right|=\ln\frac{1-\cos\frac{x}{2}}{\left|\sin\frac{x}{2}\right|}\\&=\ln\Big(\left|\csc\frac{x}{2}\right|+\left|\cot\frac{x}{2}\right|\Big)=-\ln\Big(\left|\csc\frac{x}{2}\right|-\left|\cot\frac{x}{2}\right|\Big),\end{aligned} $$

因此有

$$ \int\frac{\sqrt{1+\cos x}}{\sin x}\mathrm{d}x=\sqrt{2}\ln\left(\left|\csc\frac{x}{2}\right|-\left|\cot\frac{x}{2}\right|\right)+C. $$

(23)

$$ \int\frac{x^{3}}{\left(1+x^{8}\right)^{2}}\mathrm{d}x=\frac{1}{4}\int\frac{1}{\left(1+x^{8}\right)^{2}}\mathrm{d}\left(x^{4}\right)\xlongequal{u=x^{4}}\frac{1}{4}\int\frac{1}{\left(1+u^{2}\right)^{2}}\mathrm{d}u, $$

设 $ u = \tan t \left(-\frac{\pi}{2} < t < \frac{\pi}{2}\right) $，则 $ 1 + u^{2} = \sec^{2}t $， $ du = \sec^{2}t \, dt $，于是

$$ \begin{aligned} 原式 &=\frac{1}{4}\int\cos^{2}t\mathrm{d}t=\frac{2t+\sin2t}{16}+C\\&=\frac{\arctan x^{4}}{8}+\frac{x^{4}}{8\left(1+x^{8}\right)}+C.\end{aligned} $$

(24)

$$ \begin{aligned}\int\frac{x^{11}}{x^{8}+3x^{4}+2}\mathrm{d}x&\xlongequal{u=x^{4}}\frac{1}{4}\int\frac{u^{2}}{u^{2}+3u+2}\mathrm{d}u\\&=\frac{1}{4}\int\left(1+\frac{1}{u+1}-\frac{4}{u+2}\right)\mathrm{d}u\\&=\frac{1}{4}u+\frac{1}{4}\ln\mid1+u\mid-\ln\mid2+u\mid+C\\&=\frac{x^{4}}{4}+\ln\frac{\sqrt[4]{1+x^{4}}}{2+x^{4}}+C.\end{aligned} $$

(25)

$$ \begin{aligned}\int\frac{\mathrm{d}x}{16-x^{4}}&=\int\frac{1}{\left(2-x\right)\left(2+x\right)\left(4+x^{2}\right)}\mathrm{d}x\\&=\int\left[\frac{1}{32\left(2-x\right)}+\frac{1}{32\left(2+x\right)}+\frac{1}{8\left(4+x^{2}\right)}\right]\mathrm{d}x\\&=\frac{1}{32}\ln\left|\frac{2+x}{2-x}\right|+\frac{1}{16}\arctan\frac{x}{2}+C.\end{aligned} $$

（26）解法一

令 $ u = \tan \frac{x}{2} $，得

$$ \begin{aligned}\int\frac{\sin x}{1+\sin x}\mathrm{d}x&=\int\frac{4u}{\left(1+u\right)^{2}\left(1+u^{2}\right)}\mathrm{d}u=\int\left[\frac{-2}{\left(1+u\right)^{2}}+\frac{2}{1+u^{2}}\right]\mathrm{d}u\\&=\frac{2}{1+u}+2\arctan u+C=\frac{2}{1+\tan\frac{x}{2}}+x+C.\end{aligned} $$

解法二

$$ \begin{align*}\int\frac{\sin x}{1+\sin x}\mathrm{d}x&=\int\frac{\sin x(1-\sin x)}{\cos^{2}x}\mathrm{d}x\\&=-\int\frac{1}{\cos^{2}x}\mathrm{d}(\cos x)-\int(\sec^{2}x-1)\mathrm{d}x\\&=\sec x-\tan x+x+C.\end{align*} $$

(27)

$$ \begin{array}{l}\displaystyle\int\frac{x+\sin x}{1+\cos x}\mathrm{d}x\quad=\quad\int\frac{x}{2}\sec^{2}\frac{x}{2}\mathrm{d}x\quad+\quad\int\tan\frac{x}{2}\mathrm{d}x\\ \displaystyle\quad=\quad\int x\mathrm{d}\Bigl(\tan\frac{x}{2}\Bigr)+\quad\int\tan\frac{x}{2}\mathrm{d}x\\ \displaystyle\quad=\quad x\tan\frac{x}{2}+C.\end{array} $$

（28）

$$ \begin{aligned}\int\mathrm{e}^{\sin x}\frac{x\cos^{3}x-\sin x}{\cos^{2}x}\mathrm{d}x&=\int x\mathrm{e}^{\sin x}\cos x\mathrm{d}x-\int\mathrm{e}^{\sin x}\tan x\sec x\mathrm{d}x\\&=\int x\mathrm{d}(\mathrm{e}^{\sin x})-\int\mathrm{e}^{\sin x}\mathrm{d}(\sec x)\\&=\ x\mathrm{e}^{\sin x}-\int\mathrm{e}^{\sin x}\mathrm{d}x-(\sec x\mathrm{e}^{\sin x}-\int\mathrm{e}^{\sin x}\mathrm{d}x)\\&=\ (x-\sec x)\mathrm{e}^{\sin x}+C.\\ \end{aligned} $$

(29)

$$ \begin{align*}\int\frac{\sqrt[3]{x}}{x\left(\sqrt{x}+\sqrt[3]{x}\right)}\mathrm{d}x&\xlongequal{x=u^{6}}\int\frac{6}{u\left(u+1\right)}\mathrm{d}u=6\int\left(\frac{1}{u}-\frac{1}{u+1}\right)\mathrm{d}u\\&=6\ln\left|\frac{u}{1+u}\right|+C=\ln\frac{x}{\left(\sqrt[6]{x}+1\right)^{6}}+C.\end{align*} $$

(30)

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\left(1+\mathrm{e}^{x}\right)^{2}}&\xlongequal{x=\ln u}\int\frac{\mathrm{d}u}{u\left(1+u\right)^{2}}=\iint\left[\frac{1}{u}-\frac{1}{1+u}-\frac{1}{\left(1+u\right)^{2}}\right]\mathrm{d}u\\&=\ln u-\ln(1+u)+\frac{1}{1+u}+C\\&=x-\ln(1+\mathrm{e}^{x})+\frac{1}{1+\mathrm{e}^{x}}+C.\end{aligned} $$

(31)

$$ \begin{aligned}\int\frac{\mathrm{e}^{3x}+\mathrm{e}^{x}}{\mathrm{e}^{4x}-\mathrm{e}^{2x}+1}\mathrm{d}x&=\int\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{\mathrm{e}^{2x}-1+\mathrm{e}^{-2x}}\mathrm{d}x=\int\frac{\mathrm{d}(\mathrm{e}^{x}-\mathrm{e}^{-x})}{(\mathrm{e}^{x}-\mathrm{e}^{-x})^{2}+1}\\&=\arctan(\mathrm{e}^{x}-\mathrm{e}^{-x})+C.\end{aligned} $$

(32)

$$ \begin{align*}\int\frac{x\mathrm{e}^{x}}{\left(\mathrm{e}^{x}+1\right)^{2}}\mathrm{d}x&=-\int x\mathrm{d}\left(\frac{1}{\mathrm{e}^{x}+1}\right)=-\frac{x}{\mathrm{e}^{x}+1}+\int\frac{\mathrm{d}x}{\mathrm{e}^{x}+1}\\&=-\frac{x}{\mathrm{e}^{x}+1}+\int\frac{\mathrm{e}^{-x}\mathrm{d}x}{1+\mathrm{e}^{-x}}\\&=-\frac{x}{\mathrm{e}^{x}+1}-\ln(1+\mathrm{e}^{-x})+C.\end{align*} $$

(33)

$$ \begin{align*}\int\ln^{2}(x+\sqrt{1+x^{2}})\mathrm{d}x&=x\ln^{2}(x+\sqrt{1+x^{2}})-\int\frac{2x\ln(x+\sqrt{1+x^{2}})}{\sqrt{1+x^{2}}}\mathrm{d}x\\&=x\ln^{2}(x+\sqrt{1+x^{2}})-\int2\ln(x+\sqrt{1+x^{2}})\mathrm{d}(\sqrt{1+x^{2}})\\&=x\ln^{2}(x+\sqrt{1+x^{2}})-2\sqrt{1+x^{2}}\ln(x+\sqrt{1+x^{2}})+2x+C.\end{align*} $$

(34)

$$ \begin{aligned}\int\frac{\ln x}{\left(1+x^{2}\right)^{\frac{3}{2}}}\mathrm{d}x&\xlongequal{x=\frac{1}{u}}\int\frac{u\ln u}{\left(1+u^{2}\right)^{\frac{3}{2}}}\mathrm{d}u=-\int\ln u\mathrm{d}(\left(1+u^{2}\right)^{-\frac{1}{2}})\\&=-\frac{\ln u}{\sqrt{1+u^{2}}}+\int\frac{\mathrm{d}u}{u\sqrt{1+u^{2}}}\\&=\frac{x\ln x}{\sqrt{1+x^{2}}}-\int\frac{\mathrm{d}x}{\sqrt{1+x^{2}}}\\&=\frac{x\ln x}{\sqrt{1+x^{2}}}-\ln(x+\sqrt{1+x^{2}})+C.\end{aligned} $$

（35）设 $ x = \sin u\left(-\frac{\pi}{2} < u < \frac{\pi}{2}\right) $，则 $ \sqrt{1 - x^{2}} = \cos u, \mathrm{d}x = \cos u \, \mathrm{d}u $，于是

$$ \begin{align*}\int\sqrt{1-x^{2}}\arcsin x\mathrm{d}x&=\int u\cos^{2}u\mathrm{d}u=\frac{1}{2}\int u(1+\cos2u)\mathrm{d}u\\&=\frac{1}{4}\int u\mathrm{d}(2u+\sin2u)\\&=\frac{u(2u+\sin2u)}{4}-\frac{1}{4}\int(2u+\sin2u)\mathrm{d}u\\&=\frac{u^{2}}{4}+\frac{u}{4}\sin2u-\frac{\sin^{2}u}{4}+C\\&=\frac{(\arcsin x)^{2}}{4}+\frac{x}{2}\sqrt{1-x^{2}}\arcsin x-\frac{x^{2}}{4}+C.\end{align*} $$

（36）设 $ x = \cos u (0 < u < \pi) $，则 $ \sqrt{1 - x^{2}} = \sin u, \mathrm{d}x = -\sin u \mathrm{d}u $，于是

$$ \begin{aligned}\int\frac{x^{3}\arccos x}{\sqrt{1-x^{2}}}\mathrm{d}x&=-\int u\cos^{3}u\mathrm{d}u=-\int u\mathrm{d}\Big(\sin u-\frac{1}{3}\sin^{3}u\Big)\\&=-\ u\Big(\sin u-\frac{1}{3}\sin^{3}u\Big)+\int\Big(\sin u-\frac{1}{3}\sin^{3}u\Big)\mathrm{d}u\\&=-\ u\Big(\sin u-\frac{1}{3}\sin^{3}u\Big)-\frac{1}{3}\int(2+\cos^{2}u)\mathrm{d}(\cos u)\\&=-\ u\Big(\sin u-\frac{1}{3}\sin^{3}u\Big)-\frac{2}{3}\cos u-\frac{1}{9}\cos^{3}u+C\\&=-\frac{1}{3}\sqrt{1-x^{2}}(2+x^{2})\arccos x-\frac{1}{9}x(6+x^{2})+C.\\ \end{aligned} $$

(37)

$$ \begin{aligned}\int\frac{\cot x}{1+\sin x}\mathrm{d}x&=\int\frac{\cos x}{\sin x(1+\sin x)}\mathrm{d}x=\int\bigg(\frac{1}{\sin x}-\frac{1}{1+\sin x}\bigg)\mathrm{d}(\sin x)\\&=\left.\ln\right|\frac{\sin x}{1+\sin x}\bigg|+C.\end{aligned} $$

(38)

$$ \begin{aligned}\int\frac{\mathrm{d}x}{\sin^{3}x\cos x}&=-\int\cot x\sec^{2}x\mathrm{d}(\cot x)\xlongequal{u=\ \cot x}-\int u\bigg(1+\frac{1}{u^{2}}\bigg)\mathrm{d}u\\&=-\frac{u^{2}}{2}-\ln|\ u\ |+C\ =-\frac{\cot^{2}x}{2}-\ln|\ \cot x\ |+C.\end{aligned} $$

(39)

$$ \begin{aligned}\int\frac{\mathrm{d}x}{(2+\cos x)\sin x}&=\int\frac{\mathrm{d}(\cos x)}{(2+\cos x)(\cos^{2}x-1)}\\&\xlongequal{u=\cos x}\int\frac{\mathrm{d}u}{(2+u)(u^{2}-1)}\\&=\iint\left[\frac{1}{6(u-1)}-\frac{1}{2(u+1)}+\frac{1}{3(u+2)}\right]\mathrm{d}u\\&=\frac{1}{6}\ln|u-1|-\frac{1}{2}\ln|u+1|+\frac{1}{3}\ln|u+2|+C\\&=\frac{1}{6}\ln(1-\cos x)-\frac{1}{2}\ln(1+\cos x)+\frac{1}{3}\ln(2+\cos x)+C.\end{aligned} $$

（40）解法一

$$ \begin{aligned}\int\frac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x&=\int\frac{\frac{1}{2}(\sin x+\cos x)^{2}-\frac{1}{2}}{\sin x+\cos x}\mathrm{d}x\\&=\frac{1}{2}\int(\sin x+\cos x)\mathrm{d}x-\frac{1}{2}\int\frac{1}{\sin x+\cos x}\mathrm{d}x\\&=\frac{1}{2}(-\cos x+\sin x)-\frac{1}{2}\int\frac{1}{\sin x+\cos x}\mathrm{d}x,\end{aligned} $$

令 $ u = \tan \frac{x}{2} $，则 $ \sin x = \frac{2u}{1 + u^{2}} $， $ \cos x = \frac{1 - u^{2}}{1 + u^{2}} $， $ dx = \frac{2}{1 + u^{2}}du $，故有

$$ \begin{aligned}\int\frac{1}{\sin x+\cos x}\mathrm{d}x&=\int\frac{2}{2u+1-u^{2}}\mathrm{d}u=-\int\frac{2}{\left(u-1\right)^{2}-\left(\sqrt{2}\right)^{2}}\mathrm{d}u\\&=-\frac{1}{\sqrt{2}}\int\frac{1}{u-1-\sqrt{2}}\mathrm{d}u+\frac{1}{\sqrt{2}}\int\frac{1}{u-1+\sqrt{2}}\mathrm{d}u\\&=\frac{1}{\sqrt{2}}\ln\left|\frac{u-1+\sqrt{2}}{u-1-\sqrt{2}}\right|+C^{\prime},\end{aligned} $$

因此有

$$ \int\frac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x=\frac{1}{2}(\sin x-\cos x)-\frac{1}{2\sqrt{2}}\ln\left|\frac{\tan\frac{x}{2}-1+\sqrt{2}}{\tan\frac{x}{2}-1-\sqrt{2}}\right|+C. $$

解法二

$$ \begin{aligned}\int\frac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x&=\int\frac{\sin x\cos x}{\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)}\mathrm{d}x\xlongequal{u=x+\frac{u}{4}}\int\frac{2\sin^{2}u-1}{2\sqrt{2}\sin u}\mathrm{d}u\\&=\frac{1}{\sqrt{2}}\int\sin u\mathrm{d}u-\frac{1}{2\sqrt{2}}\int\csc u\mathrm{d}u\\&=-\frac{\cos\left(x+\frac{\pi}{4}\right)}{\sqrt{2}}-\frac{1}{2\sqrt{2}}\ln\left|\csc\left(x+\frac{\pi}{4}\right)-\cot\left(x+\frac{\pi}{4}\right)\right|+C.\end{aligned} $$