常用积分表

必背

\begin{array}{}
(1) \int kdx = kx +C & (2) \int x^adx = \frac{x^{a+1}}{a+1}+C\\
(3) \int{\frac{1}{x}}dx = \ln|x|+C& (4) \int \frac{1}{1+x^2}dx = \arctan x+C \\
(5)\int{\frac{1}{\sqrt{ 1-x^2 }}dx}=\arcsin x+C & (6)\int a^xdx=\frac{1}{\ln a}a^x+C \\
(7)\int \cos xdx = \sin x+C & (8)\int \sin xdx= -\cos x+C \\
(9)\int \sec^2xdx= \tan x+C & (10)\int \csc^2xdx=-\cot x+C \\
(11)\int \sec x\tan xdx=\sec x+C&(12)\int \csc x\cot xdx=-\csc x+C \\
(13)\int \sec xdx=\ln|\sec x+\tan x|+C&(14)\int \csc xdx=\ln|\csc x-\cot x|+C
\end{array}

考前必背

\begin{array}{#} \\
(1)\int \tan xdx=-\ln|\cos x|+C&(2)\int \cot xdx=\ln|\sin x|+C \\
(3)\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan{\frac{x}{a}}+C&(4)\int \frac{dx}{\sqrt{ x^2\pm a^2 }}=\ln|x+\sqrt{ x^2\pm a^2 }|+C \\
(5)\int \frac{dx}{\sqrt{ a^2-x^2 }}=\arcsin {\frac{x}{a}}+C&(6)\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln|{\frac{{a+x}}{a-x}}|+C
\end{array}

例题

例一 \int\frac{{1}}{x^4(x^2+1)}dx

关键点
通过分项的方法: 抄分母, 加一个, 减一个.

\begin{aligned}
\text{原式}=\int\frac{{(1+x^2)-x^2}}{x^4(x^2+1)}dx&=\int \frac{1}{x^4}dx-\int \frac{1}{x^2(x^2+1)}dx\\
&=\int\frac{{1}}{x^4}dx-\int{\frac{{1+x^2-x^2}}{x^2(x^2+1)}}dx\\
&=\int \frac{1}{x^4}dx-\int \frac{1}{x^2}dx+\int \frac{1}{x^2+1}dx\\
&=-\frac{1}{3}x^{-3}-\frac{1}{x}+\arctan x+C
\end{aligned}

例二 \int \tan^2xdx

关键点
\tan^2x+1=\sec^2x
可以由 \sin^2x+\cos^2x=1两边同时除以 \cos^2x得到

\begin{align}
原式 &=\int(\sec^2-1)dx\\ \\
&=\tan x-x+C
\end{align}

例三 \int \cos^2{\frac{x}{2}}dx

关键点:三角恒等变换
\sin\alpha \cos\alpha=\frac{1}{2}\sin2\alpha
\cos^2\alpha=\frac{{1+\cos 2\alpha}}{2}
\sin^2\alpha=\frac{{1-\cos2\alpha}}{2}

\begin{align}
\text{原式}&=\int{\frac{{1+\cos x}}{2}}dx \\
&=\int \frac{1}{2}dx+\int \frac{{\cos x}}{2}dx \\
&=\frac{1}{2}x+\frac{1}{2}\sin x+C
\end{align}

常用凑微分公式

\begin{array}{#}
\int f(ax+b) \, dx = \frac{1}{\alpha} \int f(ax+b) \, d(ax+b) (a \neq 0)\\
\int f (\sin x) \cos x \, dx = \int f (\sin x) \, d \sin x\\
\int f(\ln x) \frac{1}{x} \, dx = \int f(\ln x) \, d \ln x \, \\
\int f \left( \sqrt{x} \right) \frac{dx}{\sqrt{x}} = 2 \int r \left( \sqrt{x} \right) d\left( \sqrt{x} \right)\\
\int f (\arctan x) \frac{dx}{1 + x^2} = \int f (\arctan x) d \arctan x
\end{array}

例题

例四 \int \cos (e^x)e^xdx

关键点: 第一类换元积分法
\int f \left[ \varphi (x) \right] \varphi' (x) \, dx = \left ( \int f (u) \, du \right)_{u = \varphi (x)}

df(x)=f'(x)dx\to de^x=e^xdx
本例中可将 e^x看作 u

\begin{align}
\text{原式}&=\int \cos e^xde^x \\
&=\sin e^x +C
\end{align}

例五 \int \tan xdx

\begin{align}
\text{原式}&=\int \frac{{\sin x}}{\cos x}dx \\
&=-\int \frac{1}{\cos x}d\cos x \\
&=-\ln|\cos x|+C
\end{align}

例六 \int \frac{{1}}{\sqrt{ a^2-x^2 }}dx

\begin{align}
\text{原式}&=\int{\frac{{1}}{\sqrt{ a^2\left( 1-\left( \frac{x}{a} \right)^2 \right) }}}dx \\
&=\int {\frac{{1}}{\sqrt{ 1-\left( \frac{x}{a} \right)^2 }}}{\cdot\frac{{1}}{a}}dx \\
&=\int {\frac{{1}}{\sqrt{ 1+\left( \frac{x}{a} \right)^2 }}}d{\frac{x}{a}} \\
&=\arcsin{\frac{x}{a}}+C
\end{align}

例七 \int \frac{{dx}}{\sqrt{ x }(1+x)}

关键点
d\sqrt{ x }=(\sqrt{ x })'dx=\frac{{1}}{2\sqrt{ x }}dx
所以
{\frac{dx}{\sqrt{ x }}=2d\sqrt{ x }}

\begin{align}
\text{原式}&=2\int \frac{{1}}{1+(\sqrt{ x })^2}d\sqrt{ x } \\
&=2\arctan \sqrt{ x }
\end{align}

常见换元法

第二类换元积分法
\int f (x) \, dx = \left ( \int f \left[ \varphi (t) \right] \varphi'(t) \, dt \right)_{t = \varphi^{-1}(x)}

三角代换

被积函数含 \sqrt{ a^2-x^2 }(a>0), 令 x=a\sin t,t \in\left[ {-\frac{\pi}{2},{\frac{\pi}{2}}} \right], 则 \sqrt{ a^2-x^2 }=a\cos t.
被积函数含 \sqrt{ x^2+a^2 }(a>0), 令 x=a\tan t,t \in\left[ {-\frac{\pi}{2},{\frac{\pi}{2}}} \right], 则 \sqrt{ x^2+a^2 }=a\sec t.
被积函数含 \sqrt{ x^2-a^2 }(a>0), 令 x=a\sec t,t \in\left[ {0},{\frac{\pi}{2}})\cup({\frac{\pi}{2},\pi} \right], 则 \sqrt{ x^2-a^2 }=a\tan t.

根式代换

见例八

例题

例八 \int \frac{dx}{{1+\sqrt{ x }}}

关键点: 根式代换

\begin{align}
\text{令}t=\sqrt{ x } \\
\text{原式}&=\int {\frac{{dt^2}}{1+t}} \\
&=\int {\frac{{2tdt}}{1+t}}=2\int {\frac{{(t+1)-1}}{t+1}}dt \\
&=2\left[ \int dt-\int \frac{1}{1+t}dt \right] \\
&=2[t-\ln (t+1)]+C \\
&=2\sqrt{ x }\cdot 2\ln(\sqrt{ x }+1)+C
\end{align}

例九 \int \sqrt{ a^2-x^2 }dx

关键点: 三角代换
利用图形将新元替换为旧元

\begin{align}
\text{令}x=a\sin t \\
\text{原式}&=\int a\cos t\cdot a\cos tdt \\
&=\int a^2\cos^2tdt=a^2\int \cos^2tdt=\frac{a^2}{2}\int(1+\cos 2t)dt \\
&=\frac{a^2}{2}\left( \int dt + \frac{1}{2}\int \cos 2t d 2t\right) \\
&=\frac{a^2}{2}t+\frac{a^2}{4}\sin 2t+C
\end{align}

所以

\begin{align}
\text{原式}&=\frac{a^2}{2}\arcsin \frac{{x}}{a} +\frac{a^2}{2}\cdot \frac{x}{a}\cdot{\frac{\sqrt{ a^2-x^2 }}{a}}+C \\
&=\frac{a^2}{2}\arcsin \frac{x}{a}+\frac{x}{a}\sqrt{ a^2-x^2 }+C
\end{align}

分部积分法

\begin{align}
\int u(x)dv(x)=u(x)v(x)-\int v(x)du(x)
\end{align}

技巧
一般可依次选择 u 的顺序为: 反三角函数, 对数函数, 幂函数, 指数函数. 三角函数 (反对幂指三)

例题

例十 \int x^2\ln xdx

技巧
灵活结合换元法

\begin{align}
\text{原式}&=\int \ln x x^2dx=\frac{1}{3}\int \ln x dx^3 \\
&=\frac{1}{3}\left( \ln x\cdot x^3-\int x^3d\ln x \right) \\
&=\frac{1}{3}\left( x^3\ln x-\int x^2dx \right) \\
&=\frac{1}{3}\left( x^3\ln x-\frac{1}{3}x^3 \right)+C
\end{align}

例十一 $\int {\frac{{xe^x}}{(x+1)^2}}dx

关键
\int \frac{e^x}{(x+1)^2}dx=\int \frac{e^x}{(x+1)^2}d(x+1)

\begin{align}
\text{原式}&=\int {\frac{{(x+1)-1}}{(x+1)^2}}\cdot e^xdx=\int \frac{e^x}{x+1}dx-\int \frac{e^x}{(x+1)^2}dx \\
&=\int \frac{e^x}{x+1}dx+\int e^xd \frac{{1}}{x+1} \\
&=\int \frac{e^x}{x+1}dx+e^x\cdot \frac{{1}}{x+1}-\int \frac{e^x}{x+1}dx \\
&=\frac{e^x}{x+1}+C
\end{align}

含 \sin^mx\cdot \cos^nx的积分

\begin{align}
\sin x \text{ 奇次，} \cos x \text{ 偶次} \rightarrow d(\cos x) \text{ 或 } d(\cos x) \\
\cos x \text{ 奇次，} \sin x \text{ 偶次} \rightarrow d(\sin x) \text{ 或 } d(\cos x) \\
\cos x \text{ 和 } \sin x \text{ 同为偶次或奇次} \rightarrow d(\tan x) \text{ 或 } d(\cos x)
\end{align}

例题

例十二 \int \frac{1}{\sin^3x\cdot \cos x}dx

关键点
将 \sec^4x替换为 \sec^2x\cdot \sec^2x用后面的 \sec^2x与 dx凑微分

\begin{align}
\text{上下同时除以}\cos^4x \\
\text{原式}&=\int \frac{\sec^4x}{\tan^3x}dx \\
&=\int {\frac{{(\tan^2x+1)d\tan x}}{\tan^3x}}=\int \left( {\frac{1}{\tan x}+\tan^{-3}x} \right)d\tan x \\
&=\ln|\tan x|-\frac{1}{2}\tan^{-2}x+C
\end{align}