有理分式拆分

多项式除法（化简假分式）

假分式：分子 ≥ 分母 的分式

注意：多项式顺序的摆放，次数高的在前面，次数低的在后面，有缺陷要补齐

$$\begin{array}{r}缺x的1次项\\ eg:x^3+x^2+0x+5\end{array}$$

$$x^3+3x^2+3x+1=(x+1)(x^2+2x+1)$$

来看一下有余数的例子

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

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因式分解$x^3-3x^2+4$
先试根

$$\begin{array}{r}x=1\\ 1-3+4=2>0\\ x=-1\\ -1-3+4=0\\ \therefore x=-1为一个根\\ 可以分离出(x+1)\end{array}$$

开始多项式除法

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配凑

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$x^2(x^2+1)-x^2$

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

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

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

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

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

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通用拆分方式

有理真分式$\frac{P(x)}{Q(x)}=\frac{P(x)}{c_0(x-a{)}^k{\left(x^2+px+q\right)}^l}(k,l\text{ 为正整数},p^2-4q<0)$
可唯一地分解为有限个最简有理分式之和，即
$\frac{P(x)}{Q(x)}=\frac{A_1}{x-a}+\frac{A_2}{(x-a{)}^2}+\cdots +\frac{A_k}{(x-a{)}^k}+\frac{B_{1}x+D_1}{x^2+px+q}+\frac{B_{2}x+D_2}{(x^2+px+q{)}^2}+\cdots +\frac{B_{l}x+D_l}{(x^2+px+q{)}^l}$

$$\begin{array}{r}\frac{2x+1}{(2x+3{)}^2(2x^2+3x+4)}=\frac{A_1}{2x+3}+\frac{A_2}{(2x+3{)}^2}+\frac{Bx+D}{2x^2+3x+4}\\ \frac{2x+6x^2}{(x-5){\left(3x^2+x+1\right)}^2}=\frac{A}{x-5}+\frac{B_{1}x+D_1}{3x^2+x+1}+\frac{B_{2}x+D_2}{(3x^2+x+1{)}^2}\\ \frac{x-3}{(x-1{)}^2(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{D}{x+1}\\ \frac{3}{(x+1)(x^2-x+1)}=\frac{A}{x+1}+\frac{Bx+D}{x^2-x+1}\end{array}$$

待定系数法

$$\frac{x}{(x+1)(x+2)}$$ $$\begin{array}{r}设\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\\ \frac{A(x+2)+B(x+1)}{(x+1)(x+2)}=\frac{Ax+2A+Bx+B}{(x+1)(x+2)}\\ \therefore (A+B)x+2A+B=x\\ \therefore \{\begin{array}{l}A+B=1\\ 2A+B=0\end{array}\\ 解出A=-1,B=2\\ \frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\\ =-\frac{1}{x+1}+\frac{2}{x+2}\end{array}$$

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留数法

求谁，删谁，代谁

另一种方法:留数法
$\frac{1}{(2x+1)(x+1)}=\frac{A}{2x+1}+\frac{B}{x+1}$

$$\begin{array}{r}算A，令2x+1=0\\ 解得x=-\frac{1}{2}，代入原式\\ \frac{1}{-\frac{1}{2}+1}=2=A\\ \\ \\ 算B，令x+1=0\\ 解得x=-1，代入原式\\ \frac{1}{2\cdot (-1)+1}=-1=B\end{array}$$

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公式

$$\frac{1}{(x+a)(x+b)}$$ $$\begin{array}{r}设：\frac{1}{x+a}-\frac{1}{x+b}\\ 通分\\ \frac{x+b-x-a}{(x+a)(x+b)}=\frac{b-a}{(x+a)(x+b)}\\ \therefore \frac{1}{(x+a)(x+b)}=\frac{1}{b-a}(\frac{1}{x+a}-\frac{1}{x+b})\end{array}$$

设$y=\frac{1}{x^2-2x-8}$，求$y^{(n)}$

$$\begin{array}{r}y=\frac{1}{(x+2)(x-4)}\\ =\frac{1}{-4-2}\left(\frac{1}{x+2}-\frac{1}{x-4}\right)\\ =-\frac{1}{6}(\frac{1}{x+2}-\frac{1}{x-4})\end{array}$$

提出负数不自然就交换一下

$$\begin{array}{r}y=\frac{1}{(x+2)(x-4)}=\frac{1}{(x-4)(x+2)}\\ =\frac{1}{2-(-4)}\left(\frac{1}{x-4}-\frac{1}{x+2}\right)\\ =\frac{1}{6}\left(\frac{1}{x-4}-\frac{1}{x+2}\right)\\ \\ y^{(n)}=\frac{1}{6}\left[{\left(\frac{1}{x-4}\right)}^{(n)}-{\left(\frac{1}{x+2}\right)}^{(n)}\right]\\ =\frac{1}{6}\left[\frac{1^n\cdot (-1{)}^n\cdot n!}{(x-4{)}^{n+1}}-\frac{1^n\cdot (-1{)}^n\cdot n!}{(x+2{)}^{n+1}}\right]\\ =\frac{1}{6}(-1{)}^n\cdot n!\left(\frac{1}{(x-4{)}^{n+1}}-\frac{1}{(x+2{)}^{n+1}}\right)\end{array}$$

综合题目

$$\begin{array}{l}\int \frac{x^5+x^4-8}{x^3-x}\mathrm{d}x\\ 长除变真分式\\ \int \frac{(x^3-x)(x^2+x+1)+x^2+x-8}{x^3-x}dx\\ \int x^2+x+1+\frac{x^2+x-8}{x^3-x}dx\\ 单独处理\frac{x^2+x-8}{x^3-x}\\ \frac{x^2+x-8}{x^3-x}=\frac{x^2+x-8}{x(x-1)(x+1)}\\ 待定系数\\ \frac{A}{x}+\frac{B}{x-1}+\frac{D}{x+1}\\ 解得A=8,B=-3,D=-4\\ \int (x^2+x+1+\frac{8}{x}-\frac{3}{x-1}-\frac{4}{x+1})dx\\ =\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+x+8\ln |x|-3\ln |x-1|-4\ln |x+1|+C\end{array}$$ $$\begin{array}{r}\int \frac{x-3}{(x-1)(x^2-1)}dx\\ \int \frac{x-3}{(x-1{)}^2(x+1)}dx\\ \frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{D}{x+1}\\ 解得A=1,B=-1,D=-1\\ \int \frac{1}{x-1}-\frac{1}{(x-1{)}^2}-\frac{1}{x+1}dx\\ \ln |x-1|+\frac{1}{x-1}-\ln |x+1|+C\end{array}$$ $$\begin{array}{r}\int \frac{3}{x^3+1}dx\\ \int \frac{3}{(x+1)(x^2-x+1)}dx\\ 令\frac{A}{x+1}+\frac{Bx+D}{x^2-x+1}\\ 解得A=1,B=-1,D=2\\ \int \frac{1}{x+1}+\frac{-x+2}{x^2-x+1}dx\\ \int \frac{1}{x+1}-\frac{x-2}{x^2-x+1}dx\\ 分子=分母求导\\ (x^2-x+1{)}^{\prime }=2x-1\\ \ln |1+x|-\int \frac{\frac{1}{2}(2x-1)-\frac{3}{2}}{x^2-x+1}dx\\ -\int \frac{\frac{1}{2}(x^2-x+1{)}^{\prime }-\frac{3}{2}}{x^2-x+1}dx\\ -\frac{1}{2}\int \frac{(x^2-x+1{)}^{\prime }}{x^2-x+1}dx+\frac{3}{2}\int \frac{1}{x^2-x+1}dx\end{array}$$

接下来分别算这些积分

$$\begin{array}{r}-\frac{1}{2}\int \frac{(x^2-x+1{)}^{\prime }}{x^2-x+1}dx=-\frac{1}{2}\int \frac{1}{x^2-x+1}d(x^2-x+1)=-\frac{1}{2}\ln |x^2-x+1|\\ \frac{1}{x^2-x+1}=\frac{1}{{\left(x-\frac{1}{2}\right)}^2+\frac{3}{4}}=\frac{1}{{\left(x-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\\ \int \frac{1}{{\left(x-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}d\left(x-\frac{1}{2}\right)=\frac{2}{\sqrt{3}}\arctan \frac{\left(x-\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}+C\\ \frac{3}{2}\left[\frac{2}{\sqrt{3}}\arctan \frac{\left(x-\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}+C\right]\\ \sqrt{3}\arctan \frac{2x-1}{\sqrt{3}}+C\end{array}$$

公式

$$\begin{array}{r}\int \frac{1}{x-a}d(x-a)=\ln |x-a|+C\\ \int \frac{\mathrm{d}x}{x^2+a^2}=\frac{1}{a}\arctan \frac{x}{a}+C\left(a>0\right)\\ \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C\left(a>0\right)\\ \int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln |\frac{x-a}{x+a}|+C\left(a>0\right)\end{array}$$

比如：

$$\begin{array}{r}\int \frac{1}{x^2+3}dx=\frac{1}{\sqrt{3}}\arctan \frac{x}{\sqrt{3}}+C\end{array}$$ $$\begin{array}{l}-\int \frac{1}{2^2-t^2}dt\\ a=2,x=t\\ -\left(\frac{1}{4}\ln |\frac{2+t}{2-t}|\right)+C\\ -\frac{1}{4}\ln |\frac{2+t}{2-t}|+C\end{array}$$