0405习题总结-不定积分

1 不定积分的基本概念

例1
$$\begin{gathered} f(x)=\{\begin{array}{l}x+1,x\ge 0\\ \frac{1}{2}{e}^{-x}+\frac{1}{2},x<0\end{array} \\ 求\int f(x)dx \end{gathered}$$

$$\begin{gathered} 解：f(x)在(-\infty ,+\infty )上连续 \\ \int f(x)dx=\{\begin{array}{l}\frac{1}{2}{x}^2+x+C_1,x\ge 0，C_1为任意常数\\ -\frac{1}{2}{e}^{-x}+\frac{1}{2}x+C_2,x<0,C_2为任意常数\end{array} \\ F(x)作为f(x)的一个原函数，在x=0处连续 \\ 那么C_1=-\frac{1}{2}+C_2,令C_2=C(任意常数),则C_1=-\frac{1}{2}+C \\ \int f(x)dx=\{\begin{array}{l}\frac{1}{2}{x}^2+x-\frac{1}{2}+C,x\ge 0，C为任意常数\\ -\frac{1}{2}{e}^{-x}+\frac{1}{2}x+C,x<0\end{array} \end{gathered}$$

注意：

1. $f(x)$连续，F(x)可导。
2. $f(x)$有第一类，第二类无穷间断点，则不存在原函数。
3. $f(x)$有第二类震荡间断点时，可能存在原函数。

例2 ∫max{x3,x2,1}dx\int{max\{x^3,x^2,1\}dx}∫max{x3,x2,1}dx
解：令f(x)=max{x3,x2,1},则f(x)={x2,x≤−11,−1<x<1x3,x≥1f(x)在(−∞,+∞)上连续∫f(x)={13x3+C1,x≤−1x+C,−1<x<114x4+C2,x≥1F(x)在点x=−1出连续，所以−13+C1=−1+C,则C1=−23+CF(x)在点x=1出连续，所以14+C2=1+C,则C2=34+C∫f(x)={13x3−23+C,x≤−1x+C,−1<x<114x4+34+C,x≥1解：令f(x)=max\{x^3,x^2,1\},则\\ f(x)= \begin{cases} x^2,\quad x\le -1\\ 1,\quad -1\lt x\lt1\\ x^3,\quad x\ge1 \end{cases}\\ f(x)在(-\infty,+\infty)上连续\\ \int{f(x)}= \begin{cases} \frac{1}{3}x^3+C_1,\quad x\le -1\\ x+C,\quad -1\lt x\lt1\\ \frac{1}{4}x^4+C_2,\quad x\ge1 \end{cases}\\ F(x)在点x=-1出连续，所以-\frac{1}{3}+C_1=-1+C,则C_1=-\frac{2}{3}+C\\ F(x)在点x=1出连续，所以\frac{1}{4}+C_2=1+C,则C_2=\frac{3}{4}+C\\ \int{f(x)}= \begin{cases} \frac{1}{3}x^3-\frac{2}{3}+C,\quad x\le -1\\ x+C,\quad -1\lt x\lt1\\ \frac{1}{4}x^4+\frac{3}{4}+C,\quad x\ge1 \end{cases}\\ 解：令f(x)=max{x3,x2,1},则f(x)=⎩⎨⎧​x2,x≤−11,−1<x<1x3,x≥1​f(x)在(−∞,+∞)上连续∫f(x)=⎩⎨⎧​31​x3+C1​,x≤−1x+C,−1<x<141​x4+C2​,x≥1​F(x)在点x=−1出连续，所以−31​+C1​=−1+C,则C1​=−32​+CF(x)在点x=1出连续，所以41​+C2​=1+C,则C2​=43​+C∫f(x)=⎩⎨⎧​31​x3−32​+C,x≤−1x+C,−1<x<141​x4+43​+C,x≥1​
总结：分段函数求积分

1. 确定在分段点连续
2. 求分段积分
3. 统一常数

例3 证明：函数$\arcsin (2x-1),\arccos (1-2x),\arctan \sqrt{\frac{x}{1-x}}都是\frac{1}{\sqrt{x-x^2}}$的原函数
$$\begin{gathered} 证明：通过原函数求导函数即可 \\ [\arcsin (2x-1){]}^{{}^{\prime }}=\frac{1}{\sqrt{1-(2x-1{)}^2}}\cdot 2=\frac{1}{\sqrt{x-x^2}} \\ [\arccos (1-2x){]}^{{}^{\prime }}=-\frac{1}{\sqrt{1-(1-2x{)}^2}}\cdot -2=\frac{1}{\sqrt{x-x^2}} \\ [\arctan \sqrt{\frac{x}{1-x}}{]}^{{}^{\prime }}=\frac{1}{1+\frac{x}{1-x}}\cdot (\sqrt{\frac{x}{1-x}}{)}^{{}^{\prime }}=\frac{1}{\sqrt{x-x^2}} \end{gathered}$$

2 直接积分法-基本积分公式

例1 $\int \frac{x^4}{1+x^2}dx$

多项式除法
$$\begin{gathered} 解：\int \frac{x^4}{1+x^2}dx=\int (x^2-1+\frac{1}{1+x^2})dx \\ =\frac{1}{3}{x}^3-x+\arctan x+C \end{gathered}$$
例2 求$\int \frac{1}{x^8(1+x^2)}dx$

高次多项式与低次多项式+常数乘积
$$\begin{gathered} 解：\int \frac{1}{x^8(1+x^2)}dx=\int \frac{1-x^8+x^8}{x^8(1+x^2)}dx \\ =\int \frac{(1+x^4)(1-x^2)}{x^8}dx+\int \frac{1}{1+x^2}dx \\ =\frac{1}{x}-\frac{1}{3x^3}+\frac{1}{5x^5}-\frac{1}{7x^7}+\arctan x+C \end{gathered}$$
例3 求$\int {\tan }^{2}xdx$
$$解：\int {\tan }^{2}xdx=\int ({\sec }^{2}x-1)dx=\tan x-x+C$$
例4 求$\int \frac{1}{1+\cos 2x}dx$
$$\begin{gathered} 解：\int \frac{1}{1+\cos 2x}dx=\int \frac{1}{2{\cos }^{2}x}dx \\ =\frac{1}{2}\tan x+C \end{gathered}$$
例5 求$\int \frac{1}{{\sin }^{2}x{\cos }^{2}x}dx$
$$\begin{gathered} 解:\int \frac{1}{{\sin }^{2}x{\cos }^{2}x}dx=\int \frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx \\ =\int {\sec }^{2}xdx+\int {\csc }^{2}xdx=\tan x-\cot x+C \end{gathered}$$

3 第一换元法-凑微分形式法

∫{f[g(x)]g′(x)dx}=[∫f(t)dt]t=g(x)\int{\{f[g(x)]g^{'}(x)dx\}}=[\int{f(t)dt}]_{t=g(x)}∫{f[g(x)]g′(x)dx}=[∫f(t)dt]t=g(x)​

常见凑微分公式：

• $\int f(ax+b)dx=\frac{1}{a}\int f(ax+b)d(ax+b)$

• $\int f(e^x)e^{x}dx=\int f(e^x)d{e}^x$ $\int f(a^x)a^{x}dx=\frac{1}{\ln a}\int f(a^x)d{a}^x$

• $\int f(x^{\mu })x^{\mu -1}dx=\frac{1}{\mu }\int f(x^{\mu })d(x^{\mu })$

• $\int f(\frac{1}{x})\frac{1}{x^2}dx=-\int f(\frac{1}{x})d(\frac{1}{x})$

• $\int f(\sqrt{x})\frac{1}{\sqrt{x}}dx=2\int f(\sqrt{x})d(\sqrt{x})$

• $\int f(\ln x)\frac{1}{x}dx=\int f(\ln x)d(\ln x)\int f(x\ln x)(1+\ln x)dx=\int f(x\ln x)d(x\ln x)$

• 三角函数凑微分

  • $\int f(\sin x)\cos xdx=\int f(\sin x)d(\sin x)$
  • $\int f(\cos x)\sin xdx=-\int f(\cos x)d(\cos x)$
  • $\int f(\tan x){\sec }^{2}xdx=\int f(\tan x)d(\tan x)$
  • $\int f(\cot x){\csc }^{2}xdx=-\int f(\cot x)d(\cot x)$
  • $\int f(\sec x)\sec x\tan xdx=\int f(\sec x)d(\sec x)$
  • $\int f(\csc x)\csc x\cot xdx=-\int f(\csc x)d(\csc x)$

• 反三角函数

  • $\int f(\arctan x)\frac{1}{1+x^2}dx=\int f(\arctan x)d(\arctan x)$
  • $\int f(arccotx)\frac{1}{1+x^2}dx=-\int f(arccotx)d(arccotx)$
  • $\int f(\arcsin x)\frac{1}{\sqrt{1-x^2}}dx=\int f(\arcsin x)d(\arcsin x)$
  • $\int f(\arccos x)\frac{1}{\sqrt{1-x^2}}dx=-\int f(\arccos x)d(\arccos x)$

• $\int f(x+\frac{1}{x})(1-\frac{1}{x^2})dx=\int f(x+\frac{1}{x})d(x+\frac{1}{x})\int f(x-\frac{1}{x})(1+\frac{1}{x^2})dx=\int f(x-\frac{1}{x})d(x-\frac{1}{x})$

例1 $\int {\tan }^{5}xse{c}^{3}xdx$
$$\begin{gathered} 解：\int {\tan }^{5}xse{c}^{3}xdx=\int (se{c}^{2}x-1{)}^{2}se{c}^{2}xd(secx) \\ =\int (se{c}^{6}x-2se{c}^{4}x+se{c}^{2}x)d(secx)=\frac{1}{7}se{c}^{7}x-\frac{2}{5}se{c}^{5}x+\frac{1}{3}se{c}^{3}x+C \end{gathered}$$
例2 $\int \frac{\ln (1+x)-\ln x}{x(1+x)}dx$
$$\begin{gathered} 解：\int \frac{\ln (1+x)-\ln x}{x(1+x)}dx=-\int [\ln (1+x)-\ln x]d[\ln (1+x)-\ln x] \\ =-\frac{1}{2}[\ln (1+x)-\ln x{]}^2+C \end{gathered}$$
例3 $\int \frac{(1+2x^2)e^{x^2}}{2-3x{e}^{x^2}}dx$
$$解：\int \frac{(1+2x^2)e^{x^2}}{2-3x{e}^{x^2}}dx=-\frac{1}{3}\int \frac{1}{2-3x{e}^{x^2}}d(2-3x{e}^{x^2})=-\frac{1}{3}\ln |2-3x{e}^{x^2}|+C$$
例4 $\int \frac{e^{\arctan \sqrt{1+x}}}{(2+x)\sqrt{1+x}}dx$
$$\begin{gathered} 解：\int \frac{e^{\arctan \sqrt{1+x}}}{(2+x)\sqrt{1+x}}dx=2\int e^{\arctan \sqrt{1+x}}d(\arctan \sqrt{1+x}) \\ =2e^{\arctan \sqrt{1+x}}+C \end{gathered}$$
例5 $\int \frac{\ln \tan \frac{x}{2}}{\sin x}dx$
$$\begin{gathered} 解：\int \frac{\ln \tan \frac{x}{2}}{\sin x}dx=\int \ln \tan \frac{x}{2}d(\ln \tan \frac{x}{2}) \\ =\frac{1}{2}{\ln }^2\tan \frac{x}{2}+C \end{gathered}$$

4 第二类换元法

$\int f(x)dx=[\int [f(\varphi (t))]{\varphi }^{{}^{\prime }}(t)dt{]}_{t={\varphi }^{-1}(x)}$

• 三大换元
  1. 三角换元
     • 被积函数含有$\sqrt{a^2-x^2}$：令$x=a\sin t$
     • 被积函数含有$\sqrt{x^2+a^2}$：令$x=a\tan t$
     • 被积函数含有$\sqrt{x^2-a^2}$：令$x=a\sec t$
     • 当被积函数含有$(x^2+a^2{)}^n$也考虑$x=a\tan t$
     • 当被积函数含有$(x^2-a^2{)}^n$也考虑$x=a\sec t$
  2. 根式代换
     • 被积函数含有$\sqrt[n]{\frac{ax+b}{cx+d}}:令\sqrt[n]{\frac{ax+b}{cx+d}}=t$
     • 被积函数含有$\sqrt[n]{ax+b}:令\sqrt[n]{ax+b}=t$
     • 被积函数含有$\sqrt[n]{ax+b}，\sqrt[m]{ax+b}:令\sqrt[k]{ax+b}=t,k为m，n的最小公倍数$
  3. 倒代换
     1. 当分母方次高于分子方次，可以考虑倒代换。

例1 求$\int \frac{1}{(1+x^2{)}^2}dx$
$$\begin{gathered} 解：解法一.\int \frac{1}{(1+x^2{)}^2}dx=\int \frac{1+x^2-x^2}{(1+x^2{)}^2}dx \\ =\arctan x+\frac{1}{2}\int xd(\frac{1}{1+x^2})= \\ =\arctan x+\frac{1}{2}\frac{x}{1+x^2}-\frac{1}{2}\int (\frac{1}{1+x^2})dx \\ =\frac{1}{2}\arctan x+\frac{x}{2(1+x^2)}+C \\ 解法二.令x=\tan x \\ \int \frac{1}{(1+x^2{)}^2}dx=\int \frac{1}{(1+{\tan }^{2}x{)}^2}d\tan x \\ =\int {\cos }^{2}tdt=\frac{1}{4}\sin t\cos t-\frac{1}{2}t+C=\frac{1}{2}\arctan x+\frac{x}{2(1+x^2)}+C \end{gathered}$$
例2 求$\int \frac{1}{(a{x}^2+b{)}^n}dx,n>1,a>0$
$$\begin{gathered} 解：\int \frac{1}{(a{x}^2+b{)}^n}dx=\frac{1}{b}\int \frac{1}{(a{x}^2+b{)}^{n-1}}dx+\frac{1}{2b(n-1)}\int xd\frac{1}{(a{x}^2+b{)}^{n-1}} \\ =\frac{2bn-2b-2}{2b(n-1)}\int \frac{1}{(a{x}^2+b{)}^{n-1}}dx+\frac{x}{2b(n-1)(x^2+1{)}^{n-1}} \end{gathered}$$

例3 求$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx$​
$$\begin{gathered} 解：令\sqrt[6]{x}=t,则x=t^6,dx=6t^5 \\ \int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx=\int \frac{6t^5}{t^3+t^2}dt \\ =\int 6(t^2-t+1-\frac{1}{t+1})dt=2t^3-3t^2+6t-6\ln |t+1|+C \\ =2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\ln |x^{\frac{1}{6}}+1|+C \end{gathered}$$
例4 求$\int \frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$
$$\begin{gathered} 解：令t=\sqrt{x}+\sqrt{x+1},x=\frac{1}{4}(t-\frac{1}{t}{)}^2 \\ dt=\frac{1}{2}(t-\frac{1}{t})(1+\frac{1}{t^2})dt \\ \int \frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx=\frac{1}{2}\int \frac{t-1}{t^3}dt \\ =-\frac{1}{2t}+\frac{1}{4t^2}+C=\frac{1}{4}(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}-\sqrt{x}-2)+C \end{gathered}$$
例5求$\int \frac{1}{(1+x)\sqrt{1-x^2}}dx$
$$\begin{gathered} 解：令x=\sin t,dx=\cos t \\ \int \frac{1}{(1+x)\sqrt{1-x^2}}dx=\int \frac{1}{1+\sin t}dt \\ =\int ({\sec }^{2}t-\tan t\sec t)dt=\tan t-\sec t+C \\ =\frac{x-1}{\sqrt{1-x^2}}+C \end{gathered}$$
例6 $\int \frac{1}{x(x^6+4)}dx$
$$\begin{gathered} 解：\int \frac{1}{x(x^6+4)}dx=\frac{1}{4}\int \frac{4+x^6-x^6}{x(x^6+4)}dx \\ =\frac{1}{24}\ln \frac{x^6}{x^6+4}+C \end{gathered}$$

$\int \frac{1}{x(a{x}^n+b)}dx=\frac{1}{bn}\ln |\frac{x^n}{a{x}^n+b}|,bn\ne 0$

例7 求$\int \frac{1}{1+e^{\frac{x}{2}}+e^{\frac{x}{3}}+e^{\frac{x}{6}}}dx$
$$\begin{gathered} 解：令e^{\frac{x}{6}}=t,x=6\ln t,dx=\frac{6}{t}dt \\ 原式=\int \frac{1}{1+t^3+t^2+t}\cdot \frac{6}{t}dt \\ =\int \frac{6}{(t^2+1)(t+1)t}dt \\ \frac{6}{(t^2+1)(t+1)t}=\frac{a}{t}+\frac{b}{t+1}+\frac{cx+d}{t^2+1} \\ 计算的a=6,b=-3,c=-3,d=-3 \\ 上式=\int (\frac{6}{t}-\frac{3}{t+1}-3\frac{t+1}{t^2+1})dt \\ =6\ln |t|-3\ln |t+1|-\frac{3}{2}\ln |t^2+1|-3\arctan t+C \\ =x-3\ln |e^{\frac{x}{6}}+1|-\frac{3}{2}\ln |e^{\frac{x}{3}}+1|-3\arctan e^{\frac{x}{6}}+C \end{gathered}$$

5 分部积分求不定积分

$\int udv=vu-\int vdu$

分部积分原则：

• $\int vdu比\int udv$好求
• 分部积分选u的优先级别：反，对，幂，指，三
• 循环积分：关于于不定积分的方程

例1 求$\int x{\sin }^{2}xdx$

分析：${\sin }^{2}x$优先降幂
$$\begin{gathered} 解：\int x{\sin }^{2}xdx=\int x\cdot \frac{1-\cos 2x}{2}dx \\ =\frac{1}{4}{x}^2-\int \frac{1}{2}x\cos 2xdx=\frac{1}{4}{x}^2-\frac{1}{4}\int xd\sin 2x \\ =\frac{1}{4}{x}^2-\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+C \end{gathered}$$
例2 求$\int x^2{e}^{2x}ex$
$$\begin{gathered} 解：\int x^2{e}^{2x}dx=\frac{1}{2}\int x^{2}d{e}^{2x} \\ =\frac{1}{2}{x}^2{e}^{2x}-\frac{1}{2}\int e^{2x}d{x}^2=\frac{1}{2}{x}^2{e}^{2x}-\frac{1}{2}\int xd{e}^{2x} \\ =\frac{1}{2}{x}^2{e}^{2x}-\frac{1}{2}x{e}^{2x}+\frac{1}{4}{e}^{2x}+C \end{gathered}$$

$\int x^m{e}^{nx}dx=\frac{1}{n}{x}^m{e}^{nx}-\frac{m}{n}\int x^{m-1}{e}^{nx}dx$

例3 求$\int \frac{\arctan e^x}{e^{2x}}dx$
$$\begin{gathered} 解：\int \frac{\arctan e^x}{e^{2x}}dx=-\frac{1}{2}\int \arctan e^{x}d{e}^{-2x} \\ =-\frac{1}{2}{e}^{-2x}\arctan e^x+\frac{1}{2}\int e^x(\frac{1}{e^{2x}}-\frac{1}{1+e^{2x}})dx \\ =-\frac{\arctan e^x}{2e^{2x}}-\frac{1}{2e^x}-\frac{1}{2}\arctan e^x+C \end{gathered}$$
例4 求$\int se{c}^{3}xdx$
$$\begin{gathered} 解：\int se{c}^{3}xdx=\int secxd\tan x=\sec x\tan x-\int \tan xd\sec x \\ =\sec x\tan x-\int ({\sec }^{3}x-\sec x)dx \\ \int se{c}^{3}x=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln |\sec x+\tan x|+C \end{gathered}$$
例5 $f(x)$的一个原函数为$x\cos x$,求$\int x\cdot f^{{}^{\prime }}(x)dx$
$$\begin{gathered} 解：由已知得，\int f(x)dx=x\cos x+C \\ f(x)=(x\cos x{)}^{{}^{\prime }}=\cos x-x\sin x \\ \int x\cdot f^{{}^{\prime }}(x)dx=\int xdf(x)=xf(x)-\int f(x)dx \\ =x(\cos x-x\sin x)-x\cos x+C=-x^2\sin x+C \end{gathered}$$
例6 设$F(x)是f(x)$ 的原函数， 且当$x\ge 0时$，$f(x)\cdot F(x)=\frac{x{e}^x}{2(1+x{)}^2}$,已知$F(0)=1,F(x)>0$.求$f(x)$
$$\begin{gathered} 解：有已知得F^{{}^{\prime }}(x)=f(x) \\ f(x)\cdot F(x)=\frac{x{e}^x}{2(1+x{)}^2}两边求积分 \\ \int f(x)\cdot F(x)dx=\int \frac{x{e}^x}{2(1+x{)}^2}dx \\ {F}^2(x)=-\int x{e}^{x}d(\frac{1}{1+x})=\frac{e^x}{1+x}+C \\ 因为F(0)=1,所以C=0 \\ 因为x\ge 0时，F(x)>0,所以F(x)=\sqrt{\frac{e^x}{1+x}} \\ f(x)=F^{{}^{\prime }}(x)=(\sqrt{\frac{e^x}{1+x}}{)}^{{}^{\prime }}=\frac{1}{2}\frac{1}{\sqrt{\frac{e^x}{1+x}}}\cdot \frac{x{e}^x}{(1+x^2)} \\ =\frac{x{e}^{\frac{1}{2}}}{2(1+x{)}^{\frac{3}{2}}} \end{gathered}$$

6 表格法积分

函数$u(x),v(x),n+1阶可导$
$$\begin{gathered} \int u{v}^{(n+1)}dx反复使用分部积分可得 \\ \int u{v}^{(n+1)}dx=\int ud{v}^{(n)}=u{v}^{(n)}-\int u^{{}^{\prime }}{v}^{(n)}dx \\ =u{v}^{(n)}-u^{{}^{\prime }}{v}^{(n-1)}+\int u^{{}^{\prime \prime }}{v}^{(n-1)}dx \\ =u{v}^{(n)}-u^{{}^{\prime }}{v}^{(n-1)}+u^{{}^{\prime \prime }}{v}^{(n-2)}-\int u^{{}^{\prime \prime \prime }}{v}^{(n-2)}dx \\ =u{v}^{(n)}-u^{{}^{\prime }}{v}^{(n-1)}+u^{{}^{\prime \prime }}{v}^{(n-2)}+\cdots +(-1{)}^n{u}^{(n)}v+(-1{)}^{n+1}\int u^{(n+1)}vdx \end{gathered}$$
表格如下：

例1 求$\int (x^3+2x)\sin xdx$
$$\begin{gathered} 解：\int (x^3+2x)\sin xdx=-(x^3+2x)\cos x+(3x^2+2)\sin x+6x\cos x-6\sin x+C \\ =(3x^2-4)\sin x-(x^3-4x)\cos x+C \end{gathered}$$
例2 $\int e^{3x}\sin 2xdx$
$$\begin{gathered} 解：\int e^{3x}\sin 2xdx=-\frac{1}{2}{e}^{3x}\cos 2x+\frac{3}{4}{e}^{3x}\sin 2x-\int \frac{9}{4}{e}^{3x}\sin 2xdx \\ \int e^{3x}\sin 2xdx=-\frac{2}{13}{e}^{3x}\cos 2x+\frac{3}{13}{e}^{3x}\sin 2x+C \end{gathered}$$

7 有理函数求积分

例1 求$\int \frac{x^3+1}{x(x-1{)}^3}dx$
$$\begin{gathered} 解：\frac{x^3+1}{x(x-1{)}^3}=\frac{a}{x}+\frac{b}{(x-1{)}^3}+\frac{c}{(x-1{)}^2}+\frac{d}{x-1} \\ 通分得，a(x-1{)}^3+bx+cx(x-1)+dx(x-1{)}^2=x^3+1 \\ 计算得a=-1,b=2,c=1,d=2,所以 \\ \int \frac{x^3+1}{x(x-1{)}^3}=\int (\frac{-1}{x}+\frac{2}{x-1}+\frac{1}{(x-1{)}^2}+\frac{2}{(x-1{)}^3})dx=-\ln |x|+2\ln |x-1|-\frac{1}{x-1}-\frac{1}{(x-1{)}^2}+C \end{gathered}$$
例2 求$\int \frac{2x+2}{(x-1)(x^2+1{)}^2}dx$
$$\begin{gathered} 解：\frac{2x+2}{(x-1)(x^2+1{)}^2}=\frac{a}{x-1}+\frac{b_{1}x+c_1}{x^2+1}+\frac{b_{2}x+c_2}{(x^2+1{)}^2} \\ 通分得，a(x^2+1{)}^2+(b_{1}x+c_1)(x-1)(x^2+1)+(b_{2}x+c_2)(x-1)=2x+2 \\ 计算得，a=1,b_1=-1,c_1=-1,b_2=-2,c_2=0 \\ \int \frac{2x+2}{(x-1)(x^2+1{)}^2}=\int (\frac{1}{x-1}-\frac{x+1}{x^2+1}-\frac{2x}{(x^2+1{)}^2})dx \\ =\ln |x-1|-\frac{1}{2}\ln |x^2+1|-\arctan x+\frac{1}{x^2+1}+C \end{gathered}$$
例3 求$\int \frac{1}{8-4\sin x+7\cos x}dx$
$$\begin{gathered} 令\tan \frac{x}{2}=t,x=2\arctan t,dx=\frac{2}{1+t^2} \\ \int \frac{1}{8-4\sin x+7\cos x}dx=\int \frac{1}{8-4\frac{2t}{1+t^2}+7\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}dt \\ =\int (\frac{1}{t-5}-\frac{1}{t-3})dt=\ln |\frac{t-5}{t-3}|+C=\ln |\frac{\tan \frac{x}{2}-5}{\tan \frac{x}{2}-3}|+C \end{gathered}$$

$\int \frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx=\int \frac{A(c\sin x+d\cos x)+B(c\sin x+d\cos x{)}^{{}^{\prime }}}{c\sin x+d\cos x}dx$

例4 求$\int \frac{4\sin x+3\cos x}{2\sin x-\cos x}dx$
$$\begin{gathered} 解:原式=\int \frac{a(2\sin x-\cos x)+b(2\sin x-\cos x{)}^{{}^{\prime }}}{2\sin x-\cos x}dx \\ a=1,b=2 \\ 原式=x+\frac{2}{2\sin x-\cos x}d(2\sin x-\cos x)=x+2\ln |2\sin x-\cos x|+C \end{gathered}$$

后记

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参考：

[1]同济大学数学系.高等数学 第七版 上册[M].北京:高等教育出版社,2014.7.

[2]【梨米特】同济七版《高等数学》全程教学视频｜纯干货知识点解析，应该是全网最细｜微积分 | 高数[CP/OL].2020-04-16.p31.