Antiderivative

Anti-derivatives

原函数F(x)，则$F^{\prime }(x)=f(x)$。
我们称函数F(x)是f(x)的原函数

$$\int f(x)dx=F(x)+C$$

integral sign:$\int$（The symbol '$\int$' in English）

设F(x)与G(x)均是f(x)的原函数,则F(x)-G(x)=C

$\int f(x)dx=h(x)+c$

$$\begin{array}{r}{F}_{(x)}=\int f(x)dx=h(x)+C_1\\ G(x)=\int f(x)hx=h(x)+C_2\\ F(x)-G(x)=C_1-C_2=C\end{array}$$

两个常数相减还是一个常数。

题目（定义）

若$\int f(x)dx=x^2{e}^{2x}+C$，求$f(x)$

$$\begin{array}{r}\int f(x)dx=x^2{e}^{2x}+C=F(x)+c\\ \therefore F(x)=x^2{e}^{2x}\\ F^{\prime }(x)=f(x)\end{array}$$

若F(x)是f(x)的一个原函数，则（）成立

A. $\int f(x)\mathrm{d}x=F(x)$

B. $\int F(x)\mathrm{d}x=f(x)$

C. $\int f(x)\mathrm{d}x=F(x)+C$
正确答案为C。

D. $\int F(x)\mathrm{d}x=f(x)+C$

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设$\int f(x)dx=\sin x+C$，则$f(x)$
$f(x)=(\sin x+c{)}^{\prime }=\cos x$

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设F(x)为$e^{-x^2}$的一个原函数，求$dF(x)$
F(x)为$e^{-x^2}$的一个原函数可以得到两个信息

1. $F^{\prime }(x)=e^{-x^2}$
2. $\int e^{-x^2}dx=F(x)$
   这题

$$\begin{array}{r}dF(x)=F^{\prime }(x)dx=e^{-x^2}dx\end{array}$$

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设f(x)的一个原函数为$\frac{1}{x}$，则$f^{\prime }(x)=$

$$\begin{array}{rl}F(x)& =\frac{1}{x}\\ f(x)& =F^{\prime }(x)={\left(\frac{1}{x}\right)}^{\prime }=-\frac{1}{x^2}\\ f^{\prime }(x)& =(-\frac{1}{x^2}{)}^{\prime }=2x^{-3}\end{array}$$

不定积分性质

$$\begin{array}{rl}dy& =y^{\prime }dx\\ d\square & ={\square }^{\prime }dx\end{array}$$

下列等式正确的是
A. $[\int f(x)dx{]}^{\prime }=f(x)$
正确$[\int f(x)dx{]}^{\prime }=(F(x)+c{)}^{\prime }=f(x)$

B.$d[\int f(x)dx]=f(x)$
错误，应该为$d[\int f(x)dx]=f(x)dx$

C.$\int F^{\prime }(x)dx=f(x)$
错误，$\int F^{\prime }(x)dx=\int f(x)dx=F(x)+C$

D.$d\left[\int f(x)dx\right]=f(x)+c$
错误：$d(F(x)+C)=f(x)dx$

求$\int d\sin x$
先把这个微分算出来$d\sin x={\left(\sin x\right)}^{\prime }dx=\cos xdx$

$$\therefore 原式=\int \cos xdx=\sin x+C$$

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设 $f(x)=e^{2x}$，求$\int f^{\prime }(x)dx$
第一种方法直接算

$$\begin{array}{rl}{f}^{\prime }(x)& =e^{2x}\cdot 2=2e^{2x}\\ \therefore 原式& =\int 2e^{2x}dx\\ & 把2提出来\\ 原式& =2\int e^{2x}dx\\ \because (e^{2x}{)}^{\prime }& =2e^{2x}\\ \therefore (\frac{1}{2}{e}^{2x}{)}^{\prime }& =e^{2x}\\ 原式& =2\cdot (\frac{1}{2}{e}^{2x}+C)\\ 答案为:& e^{2x}+C\end{array}$$

第二种方法用定义：
因为

$$\begin{array}{rl}\int f(x)dx& =F(x)+C\\ \int f^{\prime }(x)dx& =f(x)+C=e^{2x}+C\end{array}$$

直接积分法

有根号（化幂函数）

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

$$\begin{array}{rl}原式& =\int x^{\frac{1}{2}}(x^2-5)dx=\int (x^{\frac{5}{2}}-5x^{\frac{1}{2}})dx\\ & =\frac{2}{7}{x}^{\frac{7}{2}}-5\cdot \frac{2}{3}\cdot x^{\frac{3}{2}}+C\end{array}$$

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$\int \frac{\sqrt[3]{x}}{\sqrt{x}}dx$

$$\begin{array}{r}\int \frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}}dx=\int x^{-\frac{1}{6}}dx\\ \frac{6}{5}{x}^{\frac{5}{6}}+C\end{array}$$

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分子分母同次方（有相同项）

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

有相同项，分子分母凑

$$\begin{array}{rl}原式& =\int \frac{(1+x^2-1)}{1+x^2}dx=\int \left(\frac{1+x^2}{1+x^2}-\frac{1}{1+x^2}\right)dx\\ & =\int 1dx-\int \frac{1}{1+x^2}dx\\ & =x-\arctan x+C\end{array}$$

分子次方 > 分母次方（降次）

假分式

凑

$\int \frac{3x^4+3x^2+1}{x^2+1}dx$

$$\begin{array}{r}\int \frac{3x^4+3x^2}{x^2+1}+\frac{1}{x^2+1}dx\\ \int \frac{3x^2\cdot (x^2+1)}{x^2+1}+\frac{1}{x^2+1}dx\\ \int 3x^2+\frac{1}{1+x^2}dx\\ 3\int x^{2}dx+\int \frac{1}{1+x^2}dx\\ 3\cdot \frac{1}{3}{x}^3+\arctan x+C\\ x^3+\arctan x+C\end{array}$$

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$\int \frac{x^4}{1+x^2}dx$
凑分子$x^2(x^2+1)-x^2$

$$\begin{array}{r}\int \frac{(1+x^2)\cdot x^2-x^2}{1+x^2}dx\\ \int \left(\frac{(1+x^2)\cdot x^2}{1+x^2}-\frac{x^2}{1+x^2}\right)dx\\ \int \left(x^2-\frac{x^2}{1+x^2}\right)dx\\ \int \left(x^2-\frac{1+x^2-1}{1+x^2}\right)dx\\ \int \left(x^2-1+\frac{1}{1+x^2}\right)dx\\ \frac{1}{3}{x}^3-x+\arctan x+C\end{array}$$

---

$\int \frac{e^{2x}-1}{e^x-1}dx$
因为$e^{2x}=(e^x{)}^2$

$$\begin{array}{rl}原式& =\int \frac{(e^x{)}^2-1^2}{e^x-1}dx\\ & =\int \frac{(e^x+1)(e^x-1)}{e^x-1}dx\\ & =\int (e^x+1)dx\\ & =e^x+x+C\end{array}$$

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$\int \frac{(x-1{)}^3}{x^2}dx$

$$\begin{array}{r}分子拆\\ (x-1{)}^3=(x-1)(x-1{)}^2\\ =(x-1)(x^2-2x+1)\\ =x^3-3x^2+3x-1\\ \int \frac{x^3-3x^2+3x-1}{x^2}dx\\ \int xdx-\int 3dx+3\int \frac{1}{x}dx-\int \frac{1}{x^2}dx\\ \frac{1}{2}{x}^2-3x+3\ln |x|+\frac{1}{x}+C\end{array}$$

---

$$\int \frac{2x^4+x^2+3}{x^2+1}dx$$

分子进行配凑

$$\begin{array}{l}{x}^4=(x^2{)}^2=(x^2+1-1{)}^2\\ [(x^2+1)-1{]}^2\\ =(x^2+1{)}^2-2(x^2+1)+1\\ \therefore 原式=\int \frac{2(x^2+1{)}^2-4(x^2+1)+2+x^2+3}{x^2+1}dx\\ \int \frac{2(x^2+1{)}^2-4(x^2+1)+x^2+1+4}{x^2+1}dx\\ \int \frac{2(x^2+1{)}^2-3(x^2+1)+4}{x^2+1}dx\\ \int 2(x^2+1)-3+\frac{4}{1+x^2}dx\\ \int 2x^2-1+\frac{4}{1+x^2}dx\\ 2\int x^{2}dx-\int dx+4\int \frac{1}{1+x^2}dx\\ 2\cdot \frac{1}{3}{x}^3-x+4\arctan x+C\end{array}$$

现在就可以进行拆分了。

另一种凑发

$$\begin{array}{l}{x}^4=(x^4-1+1)\\ [(x^2{)}^2-1^2+1]\\ (x^2+1)(x^2-1)+1\end{array}$$

接下来代入原式也可以做出来

长除法

也可以用多项式除法

$$\begin{array}{r}2x^4+x^2+3=(x^2+1)(2x^2-1)+4\\ \therefore \int \frac{(x^2+1)(2x^2-1)+4}{1+x^2}dx\\ \int 2x^2-1+\frac{4}{1+x^2}dx\end{array}$$

分子分母有有相同项，和需要将次的方法差不多

拆分

分项：抄分母，再加一个减一个
$\int \frac{1}{x^4(x^2+1)}dx$

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

---

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

$$\begin{array}{r}\frac{1}{x^2}-\frac{1}{1+x^2}=\frac{1+x^2-x^2}{x^2(1+x^2)}=\frac{1}{x^2\cdot (1+x^2)}\end{array}$$

第一种办法裂项
$\int \frac{1}{x^2}dx=\int x^{-2}dx=-x^{-1}+C=-\frac{1}{x}+C$

$$\begin{array}{r}\int \frac{1}{x^2}-\frac{1}{1+x^2}dx=-\frac{1}{x}-\arctan x+C\end{array}$$

第二种方法用上题的抄分母

$$\begin{array}{r}\int \frac{(1+x^2)-x^2}{x^2\cdot (1+x^2)}dx\\ \int \frac{1}{x^2}dx-\int \frac{1}{1+x^2}dx\\ -\frac{1}{x}-\arctan x+C\end{array}$$

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

$$\begin{array}{r}\frac{1}{x^2-a^2}=\frac{1}{(x+a)(x-a)}\\ \frac{1}{x+a}-\frac{1}{x-a}\Rightarrow \frac{x-a-(x+a)}{(x+a)(x-a)}=\frac{-2a}{(x+a)(x-a)}\\ -\frac{1}{2a}\cdot (\frac{1}{x+a}-\frac{1}{x-a})=\frac{1}{(x+a)(x-a)}\end{array}$$

提出来是负数，一般不这样提

$$\begin{array}{r}\frac{1}{x^2-a^2}=\frac{1}{(x+a)(x-a)}=\frac{1}{(x-a)(x+a)}\\ \frac{1}{x-a}-\frac{1}{x+a}\Rightarrow \frac{(x+a)-(x-a)}{(x-a)(x+a)}=\frac{2a}{(x-a)(x+a)}\\ \frac{1}{2a}(\frac{1}{x-a}-\frac{1}{x+a})=\frac{1}{(x-a)(x+a)}\end{array}$$

现在开始解题

$$\begin{array}{r}原式=\int \frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)dx\\ \frac{1}{2a}\int \frac{1}{x-a}-\frac{1}{x+a}dx\\ 根据公式\int \frac{1}{x}dx=\ln |x|+C\\ \frac{1}{2a}(\ln |x-a|-\ln |x+a|)+C\\ \frac{1}{2a}\ln |\frac{x-a}{x+a}|+C\end{array}$$

通用拆分方法

有理分式拆分

指数型

$\int a^{x}dx=\frac{a^x}{\ln a}+C$

$$\begin{array}{r}\int 3^x\cdot e^{x}dx\\ =\int (3e{)}^{x}dx\\ =\frac{(3e{)}^x}{\ln 3e}+C\end{array}$$

---

$$\int \frac{2\cdot 3^x-5\cdot 2^x}{3^x}dx$$

拆分子

$$\begin{array}{r}\int 2-5\cdot \frac{2^x}{3^x}dx\\ \int 2dx-5\int {\left(\frac{2}{3}\right)}^{x}dx\\ 2x-5\frac{{\left(\frac{2}{3}\right)}^x}{\ln \frac{2}{3}}+C\\ =2x-5\cdot \frac{(\frac{2}{3}{)}^x}{\ln 2-\ln 3}+C\end{array}$$

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三角函数型

secx 叫 正割函数（secant）
cscx = 余割函数 (cosecant)
$1+{\tan }^{2}x={\sec }^{2}x={\left(\frac{1}{\cos x}\right)}^2$

$$\begin{array}{r}\cot x=\frac{1}{\tan x},\csc x=\frac{1}{\sin x}\\ \sec x=\frac{1}{\cos x}\end{array}$$

$1+{\cot }^{2}x={\csc }^{2}x$
${\sin }^{2}x=\frac{1-\cos 2x}{2};{\cos }^{2}x=\frac{1+\cos 2x}{2}$

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$$\begin{array}{r}\because \arcsin x+\arccos x=\frac{\pi }{2}\\ \arctan x+\mathrm{arccot}x=\frac{\pi }{2}\end{array}$$ $$\begin{array}{r}\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin x+C=-\arccos x+C\\ \int \frac{1}{1+x^2}dx=\arctan x+C=-\mathrm{arccot}x+C\end{array}$$

证明

$$\begin{array}{r}\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin x+C=\frac{\pi }{2}-\arccos x+C=-\arccos x+C\\ \int \frac{1}{1+x^2}dx=\arctan x+C=\frac{\pi }{2}-\mathrm{arccot}x+C=-\mathrm{arccot}x+C\end{array}$$

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$$\begin{array}{r}\int \csc xdx=\ln |\csc x-\cot x|+C\end{array}$$ $$\begin{array}{l}\int \csc xdx=\int \frac{1}{\sin x}dx\\ \sin 2x=2\sin x\cos x\\ \int \frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}}dx\end{array}$$

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$$\begin{array}{r}\int \sec xdx=\ln |\sec x+\tan x|+C\end{array}$$ $$\begin{array}{l}\int \sec xdx=\int \frac{1}{\cos x}dx\\ 令t=\tan \frac{x}{2}\\ \frac{x}{2}=\tan t\\ x=2\tan t\\ dx=d(2\tan t)=\frac{2}{1+x^2}dt\\ \cos x={\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2}}\\ \cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{1-t^2}{1+t^2}\\ \therefore \int \frac{1}{\cos x}dx=\int \frac{1+t^2}{1-t^2}\frac{2}{1+x^2}dt\\ 2\int \frac{1}{1-t^2}dt\\ 2\int \frac{1}{(1-t)(1+t)}dt\\ 2\int \frac{1}{1-t}+\frac{1}{1+t}dt\\ \frac{1+t+1-t}{(1-t)(1+t)}=\frac{2}{(1-t)(1+t)}\\ \therefore \int \frac{1}{1-t}+\frac{1}{1+t}dt\\ \int \frac{1}{1-t}dt+\int \frac{1}{1+t}dt\\ \ln |1+t|-\ln |1-t|+C\\ \ln |\frac{1+t}{1-t}|+C\\ \ln |\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}|+C\\ 又\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\\ \frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}=\frac{\tan \frac{\pi }{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi }{4}\tan \frac{x}{2}}\\ =\tan (\frac{\pi }{4}+\frac{x}{2})\\ \sin \left(\frac{x}{2}+\frac{\pi }{4}\right)=\sin \frac{x}{2}\cos \frac{\pi }{4}+\cos \frac{x}{2}\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}(\sin \frac{x}{2}+\cos \frac{x}{2})\\ \cos \left(\frac{x}{2}+\frac{\pi }{4}\right)=\cos \frac{x}{2}\cos \frac{\pi }{4}-\sin \frac{x}{2}\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}(\cos \frac{x}{2}-\sin \frac{x}{2})\\ \therefore \tan (\frac{\pi }{4}+\frac{x}{2})=\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\\ 分子分母同乘\cos \frac{x}{2}+\sin \frac{x}{2}\\ \frac{1+\sin x}{\cos x}=\sec \mathbf{x}+\tan \mathbf{x}\end{array}$$

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$$\begin{array}{rl}& \int \frac{\cos 2x}{\cos x-\sin x}dx\\ & =\int \frac{{\cos }^{2}x-{\sin }^{2}x}{\cos x-\sin x}dx\\ & =\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{\cos x-\sin x}dx\\ & =\int \cos x+\sin xdx\\ & \sin x-\cos x+C\end{array}$$

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$$\int \frac{1}{1+\sin x}dx$$ $$\begin{array}{r}\int \frac{1-\sin x}{(1+\sin x)(1-\sin x)}dx\\ \int \frac{1-\sin x}{1-{\sin }^{2}x}dx\\ \int \frac{1-\sin x}{{\cos }^{2}x}dx\\ \int \frac{1}{{\cos }^{2}x}dx-\int \frac{\sin x}{{\cos }^2}dx\\ \int {\sec }^{2}xdx-\int \frac{\sin x}{\cos x}\cdot \frac{1}{\cos x}dx\\ \int {\sec }^{2}xdx-\int \tan x\sec xdx\\ \tan x-\sec x+C\end{array}$$

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$$\int \frac{1-\cos x}{1+\cos x}dx$$

先拆分

$$\begin{array}{r}\int \frac{(1+\cos )-2\cos x}{1+\cos x}dx\\ \int 1-\frac{2\cos x}{1+\cos x}dx\\ \int 1dx-2\int \frac{\cos x}{1+\cos x}dx\\ \int 1dx-2\int \frac{(1+\cos x)-1}{1+\cos x}dx\\ \int 1dx-2\int \left(1-\frac{1}{1+\cos x}\right)dx\\ \int dx-2\left(\int dx-\int \frac{1}{1+\cos x}dx\right)\\ \int dx-2\int dx+2\int \frac{1}{1+\cos x}dx\\ -\int dx+2\int \frac{1}{1+\cos x}dx\end{array}$$

现在来算一下这个积分$\int \frac{1}{1+\cos x}dx$
已知$\cos 2\alpha =2{\cos }^2\alpha -1$
移项$1+\cos 2\alpha =2{\cos }^2\alpha$
令$x=2\alpha ,\alpha =\frac{x}{2}$
原式得$1+\cos x=2{\cos }^2\frac{x}{2}$
代入这个积分

$$\begin{array}{r}\int \frac{1}{1+\cos x}dx\\ \int \frac{1}{2{\cos }^2\frac{x}{2}}dx\\ \int \frac{1}{2}\cdot \frac{1}{{\cos }^2\frac{x}{2}}dx\\ \frac{1}{2}\int {\sec }^2\frac{x}{2}dx\\ 令t=\frac{x}{2},x=2t\\ 原式=\frac{1}{2}\int {\sec }^{2}td2t\\ \frac{1}{2}2\int {\sec }^{2}tdt\\ \int {\sec }^{2}tdt\\ \tan t+C\\ 即\tan \frac{x}{2}+C\end{array}$$

再代入原式

$$-x+2\tan \frac{x}{2}+C$$

万能公式

推导，令$u=\tan \frac{x}{2}$

$$\begin{array}{r}\sin x=\frac{\left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)}{1}\\ =\frac{\left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}}\\ 同除{\cos }^2\frac{x}{2}\\ =\frac{\left(2\tan \frac{x}{2}\right)}{{\tan }^2\frac{x}{2}+1}=\frac{2u}{u^2+1}\end{array}$$ $$\begin{array}{r}\cos x=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}}=\frac{1-{\tan }^2\frac{x}{2}}{{\tan }^2\frac{x}{2}+1}=\frac{1-u^2}{u^2+1}\end{array}$$ $$\begin{array}{r}du=d\tan \frac{x}{2}={\sec }^2\frac{x}{2}\cdot \frac{1}{2}dx\\ \Rightarrow \\ dx=\frac{2du}{{\sec }^2\frac{x}{2}}=\frac{2du}{1+{\tan }^2\frac{x}{2}}=\frac{2du}{1+u^2}\end{array}$$

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万能代换公式，令$u=\tan \frac{x}{2}$

$$\begin{array}{r}\sin x=\frac{2u}{1+u^2},\\ \cos x=\frac{1-u^2}{1+u^2},\\ dx=\frac{2}{1+u^2}du\end{array}$$

完美适用sinx，cosx仅一次情况

$$\begin{array}{r}\int \frac{1}{1+\sin x+\cos x}dx\\ =\int \frac{1}{1+\frac{2u}{1+u^2}+\frac{1-u^2}{1+u^2}}\cdot \frac{2}{1+u^2}du\\ =\int \frac{\frac{2}{1+u^2}}{1+\frac{2u}{1+u^2}+\frac{1-u^2}{1+u^2}}du\\ \int \frac{2}{1+u^2+2u+1-u^2}du\\ \int \frac{2}{1+2u+1}du\\ \int \frac{1}{1+u}du\\ \int \frac{1}{1+u}d(1+u)\\ \ln |1+\tan \frac{x}{2}|+C\end{array}$$

凑微分法

第一类换元积分法

更精准地说，不定积分不是求导的逆运算，而是微分的逆运算；不定积分和微分才是真正的一对。

使用条件：

• 遇到被积函数复杂
  对于一个函数$Q(x)$,它$\int Q(x)dx$求积分大致思想：

1. 拆这个函数Q(x)，$\int f[g(x)]\cdot g^{\prime }(x)dx$
2. 现在$g^{\prime }(x)dx$是一个微分
3. 所以变为$\int f[g(x)]dg(x)$
4. 令g(x)=u,$\int f(u)du$
5. $F(u)+C$
6. 最后再回代回来$F(g(x))+C$

凑微分的条件：

• 若被积函数可分成$f(x)\cdot g(x)$或$\frac{f(x)}{g(x)}$
• 且若f(x)较复杂，且f(x)求导可以得到g(x)的倍数

$\int \cos (e^x)\cdot e^{x}dx$
可以直接看出来$[\sin (e^x){]}^{\prime }=\cos (e^x)\cdot e^x$

现在用换元法

$$\begin{array}{rl}df(x)& =f^{\prime }(x)dx\\ d{e}^x& =(e^x{)}^{\prime }dx=e^{x}dx\\ \therefore 原式& =\int \cos (e^x)d(e^x)\\ 令e^x& =u\\ 原式& =\int \cos udu\\ & =\sin u+C\\ \therefore 原式& =\sin e^x+C\end{array}$$

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$$\int \frac{e^{\arccos x}}{\sqrt{1-x^2}}dx$$ $$\begin{array}{rl}& \int e^{\arccos x}\cdot \frac{1}{\sqrt{1-x^2}}dx\\ & \because (\arccos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}\\ & \therefore 原式=-\int e^{\arccos x}\cdot (\arccos x{)}^{\prime }dx\\ & -\int e^{\arccos x}d(\arccos x)\\ & 令\arccos x=u\\ & 原式=-\int e^{u}du\\ & =-e^u+C\\ & =-e^{\arccos x}+C\end{array}$$

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求$\int \frac{1}{x^2}\cdot e^{1-\frac{1}{x}}dx$
$(x^{-1}{)}^{\prime }=-x^{-2}=-\frac{1}{x^2}$
那么$\frac{1}{x^2}=-(x^{-1}{)}^{\prime }$

$$\begin{array}{r}原式=\int e^{1-\frac{1}{x}}-(x^{-1}{)}^{\prime }dx\\ \int e^{1-\frac{1}{x}}{\left(-\frac{1}{x}\right)}^{\prime }dx\\ 常数求导为0，所以再添一个1\\ \int e^{1-\frac{1}{x}}{\left(1-\frac{1}{x}\right)}^{\prime }dx\\ \int e^{1-\frac{1}{x}}d\left(1-\frac{1}{x}\right)\\ \int e^{u}du\\ e^u+C\\ e^{1-\frac{1}{x}}+C\end{array}$$

另一种解法：$\int \frac{1}{x^2}\cdot e^{1-\frac{1}{x}}dx$
复杂：$e^{1-\frac{1}{x}}$
简单：$\frac{1}{x^2}$
${\left(1-\frac{1}{x}\right)}^{\prime }=-{\left(\frac{1}{x}\right)}^{\prime }=-(-1)x^{-2}=\frac{1}{x^2}$

$$\begin{array}{r}原式=\int e^{1-\frac{1}{x}}{\left(1-\frac{1}{x}\right)}^{\prime }dx\\ \int e^{1-\frac{1}{x}}d\left(1-\frac{1}{x}\right)\end{array}$$

再换元就可以求出答案

---

$\int \frac{2x-1}{x^2-x-2}dx$

$$\begin{array}{r}\int \frac{1}{x^2-x-2}\cdot (2x-1)dx\end{array}$$

复杂：$\frac{1}{x^2-x-2}$
简单：$2x-1$
$(x^2-x-2{)}^{\prime }=2x-1$

$$\begin{array}{r}原式=\int \frac{1}{x^2-x-2}\cdot (x^2-x-2{)}^{\prime }dx\\ \int \frac{1}{u}\cdot du\\ \ln |u|+C\\ \ln |x^2-x-2|+C\end{array}$$

---

$$x{e}^{-2x^2}dx=d()$$

我们知道：$d\square ={\square }^{\prime }dx$
现在${\square }^{\prime }$已知，转换成也就是求

$$\int x{e}^{-2x^2}dx$$

复杂：$e^{-2x^2}$
简单：x
$(-2x^2{)}^{\prime }=-2\cdot 2x=-4x$
现在凑出-4x

$$\begin{array}{r}\int e^{-2x^2}(-4x)\cdot \left(-\frac{1}{4}\right)dx\\ -\frac{1}{4}\int e^{-2x^2}(-2x^2{)}^{\prime }dx\\ -\frac{1}{4}\int e^{-2x^2}d(-2x^2)\\ -\frac{1}{4}\int e^{u}d(u)\\ -\frac{1}{4}{e}^u+C\\ -\frac{1}{4}{e}^{-2x^2}+C\end{array}$$

---

若$\int f(x)\mathrm{d}x=F(x)+C$，求证明$\int e^{x}f(e^x)\mathrm{d}x=F(e^x)+C$
令$e^x=u$，得$\int uf(u)dx$

$u^{\prime }=u=e^x$

$$\begin{array}{rl}原式& =\int f(u)udx\\ & =\int f(u)u^{\prime }dx\\ & =\int f(u)du\\ & =F(u)+C\\ & =F(e^x{)}_C\end{array}$$

凑微分法 2

若被积函数中，复杂函数求导，无法得到简单函数的倍数。
考虑“被积函数分子分母”同乘或同除某因子，方便积分

$$\int \frac{1}{e^x+e^{-x}}dx$$

化简（被积函数分子分母）

$$\begin{array}{r}\int \frac{e^x}{(e^x+e^{-x})e^x}dx\\ \int \frac{e^x}{(e^x{)}^2+1}dx\\ \int \frac{1}{(e^x{)}^2+1}{e}^{x}dx\\ \int \frac{1}{(e^x{)}^2+1}d{e}^x\\ \int \frac{1}{1+(e^x{)}^2}d{e}^x\\ \int \frac{1}{1+(u{)}^2}du\\ \arctan u+C\\ \arctan (e^x)+C\end{array}$$

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凑微分法 3 三角函数次方

遇到${\sin }^{偶}x,{\sin }^{奇}x\Rightarrow 奇拆，偶降$

$$\int {\cos }^3{x}^{\cdot }dx$$ $$\begin{array}{r}=\int {\cos }^{2}x\cos xdx\\ =\int {\cos }^{2}xd\sin x\\ =\int (1-{\sin }^{2}x)d\sin x\\ =\int (1-t^2)dt\\ =\int 1dt-\int t^{2}dt\\ =t-\frac{1}{3}{t}^3+C\\ \sin x-\frac{1}{3}{\sin }^{3}x+C\end{array}$$

---

$\int \sqrt{\sin x}{\cos }^{5}xdx$

$$\begin{array}{r}\int \sqrt{\sin x}{\cos }^{4}x\cos xdx\\ \int \sqrt{\sin x}{\cos }^{4}d\sin x\\ \int \sqrt{\sin x}({\cos }^{2}x{)}^{2}d\sin x\\ \int \sqrt{\sin x}(1-{\sin }^{2}x{)}^{2}d\sin x\\ \int \sqrt{u}(1-u^2{)}^{2}du\\ \int u^{\frac{1}{2}}(1-2u^2+u^4)du\\ \int \left(u^{\frac{1}{2}}-2u^{\frac{5}{2}}+u^{\frac{9}{2}}\right)du\\ \frac{2}{3}{u}^{\frac{3}{2}}-2\cdot \frac{2}{7}{u}^{\frac{7}{2}}+\frac{2}{11}{u}^{\frac{11}{2}}+C\\ \frac{2}{3}{u}^{\frac{3}{2}}-\frac{4}{7}{u}^{\frac{7}{2}}+\frac{2}{11}{u}^{\frac{11}{2}}+C\\ \frac{2}{3}\sin x^{\frac{3}{2}}-\frac{4}{7}\sin x^{\frac{7}{2}}+\frac{2}{11}\sin x^{\frac{11}{2}}+C\end{array}$$

---

更普遍的情况：
对于含有${\sin }^{m}x\cdot {\cos }^{n}x$的积分

1. $\sin x$奇次，$\cos x$偶次，凑$d(\cos x),\text{or}d(\sec x)$
2. $\cos x$奇次，$\sin x$偶次，凑$d(\sin x),\text{or}d(\csc x)$
3. $\cos x$和$\sin x$同为偶次或奇次，凑$d(\tan x)$或$d(\cot x)$

凑微分公式法

$$\begin{array}{r}\int f(ax+b)dx\\ =\frac{1}{a}\int f(ax+b)d(ax+b)\end{array}$$

证明：$d(ax+b)=adx$

$$\begin{array}{r}原式=\frac{1}{a}\int f(ax+b)adx\\ =\int f(ax+b)dx\end{array}$$ $$\int (3x+1{)}^{2}dx$$

不用公式做：

$$\begin{array}{r}\int (9x^2+1+6x)dx\\ \int 9x^{2}dx+\int dx+\int 6xdx\\ 9\int x^{2}dx+\int dx+6\int xdx\\ 9\cdot \frac{1}{3}{x}^3+x+6\cdot \frac{1}{2}{x}^2+C\\ 3x^3+x+3x^2+C\end{array}$$

用公式

$$\begin{array}{r}\int (3x+1{)}^{2}dx=\frac{1}{3}\int (3x+1{)}^{2}d(3x+1)\\ \frac{1}{3}\int u^{2}du\\ \frac{1}{3}\cdot \frac{1}{3}{u}^3+C\\ \frac{1}{9}(3x+1{)}^3+C\\ \frac{1}{9}(27x^3+3\cdot 9x^2\cdot 1+3\cdot 3x+1)+C\\ 3x^3+3x^2+x+C\end{array}$$

$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$

---

$\int e^{5t}dt$

$$\begin{array}{r}\frac{1}{5}\int e^{5t}d5t\\ \frac{1}{5}\int e^{u}du\\ \frac{1}{5}{e}^u+C\\ \frac{1}{5}{e}^{5t}+C\end{array}$$

---

$\int e^{x-3}dx$

$$\begin{array}{r}\int e^{x-3}d(x-3)\\ \int e^{u}du\\ e^u+C\\ e^{x-3}+C\end{array}$$

第二种方法

$$\begin{array}{r}\int e^x\cdot e^{-3}dx\\ e^{-3}\int e^{x}dx\\ e^{-3}(e^x+C)\\ e^{x-3}+C\end{array}$$

---

$$\begin{array}{r}\int x^{a}dx=\frac{1}{a+1}{x}^{a+1}+C(x\ne -1)\\ \int \frac{1}{x}dx=\ln |x|+C(x\ne 0)\end{array}$$

不要混淆

$$\begin{array}{r}\int \frac{1}{1+x^2}dx=\int (1+x^2{)}^{-1}dx=\ln |1+x^2|+C\\ \int \frac{1}{x^2}dx=\int x^{-2}dx=-x^{-1}+C\end{array}$$

---

$$\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln |\frac{a+x}{a-x}|+C$$

推导过程

$$\begin{array}{r}\int \frac{1}{a^2-x^2}dx=\int \frac{1}{(a+x)(a-x)}dx\\ \frac{1}{2a}\int \frac{(a+x)+(a-x)}{(a+x)(a-x)}dx\\ \frac{1}{2a}\int \frac{1}{a-x}+\frac{1}{a+x}dx\\ \frac{1}{2a}\left(\int \frac{1}{a-x}dx+\int \frac{1}{a+x}dx\right)\\ 先求\int \frac{1}{a-x}dx\\ 令t=a-x,x=a-t\\ \int \frac{1}{t}d(a-t)=-\int \frac{1}{t}dt\\ =-\ln |t|+C=-\ln |a-x|+C\\ \therefore 原式=\frac{1}{2a}(-\ln |a-x|+\ln |a+x|+C)\\ =\frac{1}{2a}\ln |\frac{a+x}{a-x}|+C\end{array}$$ $$\begin{array}{l}\int \frac{\sin x}{4-{\cos }^{2}x}dx\\ \int \frac{1}{4-{\cos }^{2}x}d(-\cos x{)}^{\prime }\\ -\int \frac{1}{4-{\cos }^{2}x}d\cos x\\ 令t=\cos x\\ -\int \frac{1}{2^2-t^2}dt\\ a=2,x=t\\ -\left(\frac{1}{4}\ln |\frac{2+t}{2-t}|\right)+C\\ -\frac{1}{4}\ln |\frac{2+t}{2-t}|+C\\ -\frac{1}{4}\ln |\frac{2+\cos x}{2-\cos x}|+C\end{array}$$

常见微分导数公式

$$\begin{array}{r}\int \frac{1}{x^2}dx=-\frac{1}{x}+C\\ \int \frac{1}{2\sqrt{x}}dx=\sqrt{x}+C\end{array}$$

证明：

$$\begin{array}{r}{\left(\frac{1}{x}\right)}^{\prime }=\frac{-1}{x^2}=-\frac{1}{x^2}\\ (\sqrt{x}{)}^{\prime }={\left(x^{\frac{1}{2}}\right)}^{\prime }=\frac{1}{2}{x}^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\end{array}$$

第二类换元积分法 简单无理根式换元法

遇到$\sqrt[n]{ax+b}$
运用无理根式换元

1. 令$t=\sqrt[n]{ax+b}$
2. 反解出x
3. 求出dx
4. 回代计算

$\int x\sqrt[5]{3x+2}dx$
令$t=\sqrt[5]{3x+2}$

$$\begin{array}{r}{t}^5=3x+2\\ x=\frac{t^5-2}{3}\end{array}$$ $$\begin{array}{r}dx=d\left(\frac{t^5-2}{3}\right)\\ d\left(\frac{t^5-2}{3}\right)={\left(\frac{t^5-2}{3}\right)}^{\prime }dt\\ ={\left(\frac{1}{3}(t^5-2)\right)}^{\prime }dt={\left(\frac{1}{3}{t}^5-\frac{2}{3}\right)}^{\prime }dt\\ =\frac{1}{3}5t^{4}dt=\frac{5}{3}{t}^{4}dt\end{array}$$ $$\begin{array}{r}\therefore 原式=\int \frac{t^5-2}{3}t\cdot \frac{5}{3}{t}^{4}dt\\ \frac{1}{3}\int (t^5-2)t^5\frac{5}{3}dt\\ \frac{5}{9}\int (t^{10}-2t^5)dt\\ \frac{5}{9}\left(\frac{t^{11}}{11}-2\cdot \frac{t^6}{6}\right)+C\\ \frac{5}{9}\left(\frac{t^{11}}{11}-\frac{t^6}{3}\right)+C\\ \frac{5t^{11}}{99}-\frac{5t^6}{27}+C\\ 回代t得到答案\\ \frac{5}{99}{\left[(3x+2{)}^{\frac{1}{5}}\right]}^{11}-\frac{5}{27}{\left[(3x+2{)}^{\frac{1}{5}}\right]}^6+C\end{array}$$

---

$\int \frac{1}{1+\sqrt{x}}dx$
令$\sqrt{x}=t$,$x=t^2$

$$\begin{array}{r}dx=d{t}^2=2tdt\end{array}$$ $$\begin{array}{r}\therefore 原式=\int \frac{1}{1+t}2tdt\\ 2\int \frac{1}{1+t}tdt\\ 2\int \frac{t}{1+t}dt\\ 2\int \frac{1+t-1}{1+t}dt\\ 2\int \frac{1+t}{1+t}-\frac{1}{1+t}dt\\ 2\left(\int 1dt-\int \frac{1}{1+t}dt\right)\\ 2\left(\int 1dt-\int \frac{1}{1+t}d(1+t)\right)\\ 2(t-\ln |1+t|)+C\\ 最后回代\\ 2\sqrt{x}-2\ln |1+\sqrt{x}|+C\end{array}$$

三角代换

利用三角函数平方和公式：去掉含$x^2$的根式
条件：含$x^2$的根式$\sqrt{a+x^2}$

1. 含$\sqrt{a^2+x^2}$，令$x=a\tan t$
2. 含$\sqrt{a^2-x^2}$，令$x=a\sin t$
3. 含$\sqrt{x^2-a^2}$，令$a=\sec t$

(2)推导

$$\begin{array}{r}\sqrt{a^2-x^2}=\sqrt{a^2-a^2{\sin }^{2}t}=\sqrt{a^2(1-{\sin }^{2}t)}\\ \sqrt{a^2{\cos }^{2}t}=a\cos t\end{array}$$

---

$$\int \frac{dx}{(1-x^2{)}^{\frac{3}{2}}}$$

这也是一个根式：

$$\begin{array}{r}\int \frac{dx}{\sqrt{(1-x^2{)}^3}}\end{array}$$

令x=sint

$$\begin{array}{r}\int \frac{d\sin t}{(1-{\sin }^{2}t{)}^{\frac{3}{2}}}\\ \int \frac{\cos tdt}{({\cos }^{2}t{)}^{\frac{3}{2}}}\\ \int \frac{\cos tdt}{(\cos t^2{)}^{\frac{3}{2}}}\\ \int \frac{\cos tdt}{({\cos }^{3}t)}\\ \int \frac{1}{{\cos }^{2}t}dt\\ \int {\sec }^{2}tdt\\ \tan t+C\end{array}$$

已知：x=sint

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代回x：

$$\therefore 原式=\frac{x}{\sqrt{1-x^2}}+C$$

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$${\int }_0^1\sqrt{1-x^2}dx$$

纯用公式做，不用几何法。

$$\begin{array}{l}\\ 先求\int \sqrt{1-x^2}dx\\ 令x=\sin t,-1\le \sin t\le 1\\ \sqrt{1-{\sin }^{2}t}=\cos t\\ \int \cos td\sin t\\ \int {\cos }^{2}tdt\\ \int \frac{1+\cos 2t}{2}dt\\ 现在求\frac{1}{2}\int 1+\cos 2tdt\\ \frac{1}{2}\left(t+\frac{1}{2}\sin 2t\right)+C\\ 又\because \sin 2t=2\sin t\cos t,x=\sin t,\cos t=\sqrt{1-{\sin }^{2}t}\\ \therefore \sin 2t=2x\sqrt{1-x^2}\\ \therefore \frac{1}{2}\left(t+\frac{1}{2}\sin 2t\right)+C=\\ \frac{1}{2}\left(\arcsin x+\frac{1}{2}2x\sqrt{1-x^2}\right)+C\\ \left(\frac{1}{2}\left(\arcsin x+x\sqrt{1-x^2}\right)\right){|}_0^1\\ \frac{1}{2}\arcsin 1+0-\frac{1}{2}(\arcsin 0+0)=\frac{1}{2}\frac{\pi }{2}=\frac{\pi }{4}\end{array}$$

---

$$\begin{array}{r}\int \frac{1}{\sqrt{4-x^2}}dx\\ 令x=2\sin t,dx=d(2\sin t)=2\cos tdt\\ \int \frac{2\cos t}{\sqrt{4-4{\sin }^{2}t}}dt\\ \int \frac{2\cos t}{\sqrt{4(1-{\sin }^{2}t)}}dt\\ \int \frac{2\cos t}{2\cos t}dt\\ \int dt=t+C\\ x=2\sin t,\frac{x}{2}=\sin t\\ \therefore t=\arcsin \frac{x}{2}\\ =\arcsin \frac{x}{2}+C\end{array}$$

---

$$\begin{array}{l}{\int }_1^{\sqrt{3}}\frac{1}{x^2\sqrt{1+x^2}}dx\\ 令x=\tan t,dx=d(\tan t)={\sec }^{2}tdt\\ x=\sqrt{3},t=\frac{\pi }{3}\\ x=1,t=\frac{\pi }{4}\\ \sqrt{1+x^2}=\sqrt{1+{\tan }^{2}t}=\sqrt{{\sec }^{2}t}=\sec t\\ \therefore {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{1}{{\tan }^{2}t\sec t}\cdot {\sec }^{2}tdt\\ {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{\sec t}{{\tan }^{2}t}dt\\ {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{\left(\frac{1}{\cos t}\right)}{\frac{{\sin }^{2}t}{{\cos }^{2}t}}dt\\ {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{1}{\frac{{\sin }^{2}t}{\cos t}}dt\\ {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{\cos t}{{\sin }^{2}t}dt\\ {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{1}{{\sin }^{2}t}d(\sin t)\\ 令u=\sin t\\ t=\frac{\pi }{3},u=\frac{\sqrt{3}}{2}\\ t=\frac{\pi }{4},u=\frac{\sqrt{2}}{2}\\ {\int }_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{u}^{-2}du\\ -\frac{1}{u}{|}_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{2}}\\ =-\frac{2\sqrt{3}}{3}+\sqrt{2}\end{array}$$

分部积分

来源（导数的乘法法则）：$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$
移向

$$u{v}^{\prime }={\left(uv\right)}^{\prime }-u^{\prime }v$$

等式两边积分

$$\begin{array}{r}\Rightarrow (2):\int u{v}^{\prime }dx=\int {\left(uv\right)}^{\prime }dx-\int u^{\prime }vdx\\ \Rightarrow \int udv=\int duv-\int vdu\\ 或者\int udv=uv-\int vdu\end{array}$$

得到分部积分公式：

$$\int udv=uv-\int vdu$$

也可以用积分乘法公式推得。

由 2 式变形 得

$$\begin{array}{r}\int udv=uv-\int v{u}^{\prime }dx\end{array}$$ $$使用条件\{\begin{array}{l}1.乘积关系\\ 2.不具有导数关系\end{array}$$

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类型一

积分式子中，只有一个函数（反三角，对数）

$$\int \ln xdx$$

套公式：$\int udv=uv-\int v{u}^{\prime }dx$
u=lnx, v=x

$$\begin{array}{r}原式=\ln x\cdot x-\int x(\ln x{)}^{\prime }dx\\ \ln x\cdot x-\int x\cdot \frac{1}{x}dx\\ \ln x\cdot x-\int dx\\ \ln x\cdot x-x+C\end{array}$$

---

$$\int \arcsin xdx$$

套公式：$\int udv=uv-\int v{u}^{\prime }dx$

$$\begin{array}{rl}& 1.u=\arcsin x\\ & 2.v=x\end{array}$$ $$\begin{array}{r}原式=\arcsin x\cdot x-\int x(\arcsin x{)}^{\prime }dx\\ =\arcsin x\cdot x-\int x\frac{1}{\sqrt{1-x^2}}dx\\ \because (1-x^2{)}^{\prime }=-2x\\ 原式=\arcsin x\cdot x-\int \left(-\frac{1}{2}\right)(1-x^2{)}^{\prime }\frac{1}{\sqrt{1-x^2}}dx\\ =\arcsin x\cdot x+\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}(1-x^2{)}^{\prime }dx\\ =\arcsin x\cdot x+\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}d(1-x^2)\\ =\arcsin x\cdot x+\frac{1}{2}\int \frac{1}{\sqrt{u}}d(u)\\ \because \int \frac{1}{\sqrt{x}}dx=2\sqrt{x}+C\\ =\arcsin x\cdot x+\frac{1}{2}\cdot 2\cdot \sqrt{u}+C\\ =\arcsin x\cdot x+\sqrt{u}+C\\ =\arcsin x\cdot x+\sqrt{1-x^2}+C\end{array}$$

类型二

积分式子中两个不同类型函数相乘

$$\int x{\tan }^{2}xdx$$

两种方式：

$$\begin{array}{r}u=x\\ v^{\prime }={\tan }^{2}x\\ v=\int {\tan }^{2}xdx\\ =\int ({\sec }^2-1)dx\\ =\tan x-x+C\\ 原式=\int x({\sec }^2-1)dx\\ =\int xd(\tan x-x)\\ x(\tan x-x)-\int (\tan x-x)\cdot 1\cdot dx\\ x(\tan x-x)-\left(\int \tan xdx-\int xdx\right)\\ x(\tan x-x)-\left(-\ln |\cos x|-\frac{1}{2}{x}^2\right)+C\\ x(\tan x-x)+\ln |\cos x|+\frac{1}{2}{x}^2+C\\ x\tan x+\ln |\cos x|-\frac{1}{2}{x}^2+C\end{array}$$

第二种解法：

$$\begin{array}{r}\int x{\tan }^{2}xdx\\ 原式=\int x({\sec }^{2}x-1)dx\\ \int (x{\sec }^{2}x-x)dx\\ \int x{\sec }^{2}x-\int xdx\\ \int xd(\tan x)-\frac{1}{2}{x}^2\\ x\tan x-\int \tan xdx-\frac{1}{2}{x}^2\\ x\tan x-(-\ln |\cos x|)-\frac{1}{2}{x}^2+C\\ x\tan x+\ln |\cos x|-\frac{1}{2}{x}^2+C\end{array}$$

类型三

积分式子 $e^x,\cos x,\sin x$ 乘积型式（会循环）

$$\begin{array}{r}\int e^x\cos xdx\end{array}$$

反对幂指三

$$\begin{array}{rl}1.& u=e^x\\ 2.& v^{\prime }=\cos x\Rightarrow v=\sin x\end{array}$$ $$\begin{array}{r}原式=\int e^{x}d\sin x\\ e^x\sin x-\int \sin x{e}^{x}dx\\ e^x\sin x-\int e^{x}d(-\cos x)\\ e^x\sin x+\int e^{x}d\cos x\\ e^x\sin x+e^x\cos x-\int \cos x{e}^{x}dx\\ \therefore \int e^x\cos xdx=e^x\sin x+e^x\cos x-\int \cos x{e}^{x}dx\\ 2\int e^x\cos xdx=e^x\sin x+e^x\cos x+C\\ \therefore \int e^x\cos xdx=\frac{e^x\sin x+e^x\cos x}{2}+C\end{array}$$

类型四

积分式子中，含$f^{\prime }(x),f^{\prime \prime }(x)$抽象函数，也用分部积分（优先使用）

已知f(x)的一个原函数为$\sin x\ln x$，求$\int x{f}^{\prime }(x)dx$

$$\begin{array}{rl}(\sin x\ln x+C{)}^{\prime }& =f(x)(这题C可以不加)\\ \int f(x)dx& =\sin x\ln x+C\end{array}$$ $$\begin{array}{r}\int x{f}^{\prime }(x)dx\\ =\int xdf(x)\\ =xf(x)-\int f(x)x^{\prime }dx\\ =xf(x)-\int f(x)dx\\ x(\sin x\ln x{)}^{\prime }-\sin x\ln x+C\\ x\left(\cos x\ln x+\sin x\frac{1}{x}\right)-\sin x\ln x+C\\ x\cos x\ln x+\sin x-\sin x\ln x+C\end{array}$$

表格法

适用题型

$$\int P_n(x)e^{ax}dx,\int Q_n(x)\sin bxdx(P_n(x),Q_n(x)为多项式)$$

方法：对$P_n(x)$求导至0，对$e^{ax},\sin bx$积分

$$\begin{array}{r}\int (x^2+2x+3)e^{2x}dx\end{array}$$

设${\int }_0^{a}x{e}^{2x}dx=\frac{1}{4}$，则$a=$

导	x(+)	1(-)	0
积	$e^{2x}$	$\frac{1}{2}{e}^{2x}$	$\frac{1}{4}{e}^{2x}$

$$\begin{array}{r}\left(\frac{1}{2}{e}^{2x}\cdot x-\frac{1}{4}{e}^{2x}\right){|}_0^a=\frac{1}{4}\\ \frac{a}{2}{e}^{2a}-\frac{1}{4}{e}^{2a}-\left(-\frac{1}{4}\right)=\frac{1}{4}\end{array}$$

循环

遇到循环（导与积到原积分重现为止）

$$\int e^{2x}\cos 3xdx$$

证明题

积分带字母