微积分学/不定积分/部分积分法

维基百科中的相关条目：

部分分式积分法

例题

$\begin{array}{cc}& 1.\int \frac{x^3+4x+3}{x^4+5x^2+4}dx\\ & =\int \frac{x^3+4x+3}{(x^2+1)(x^2+4)}dx\\ & =\int \frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}dx\\ & \frac{x^3+4x+3}{(x^2+1)(x^2+4)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}\\ & \frac{x^3+4x+3}{(x^2+4)}=Ax+B+\frac{Cx+D}{x^2+4}(x^2+1)\\ & x=i\\ & \frac{-i+4i+3}{3}=i+1=Ai+B\\ & A=1B=1\\ & \frac{x^4+4x^2+3x}{x^4+5x^2+4}=\frac{x^2+x}{x^2+1}+\frac{C{x}^2+Dx}{x^2+4}\\ & x\to \mathrm{\infty }\\ & 1=1+CC=0\\ & x=0\\ & \frac{3}{4}=1+\frac{D}{4}D=-1\\ & \int \frac{x^3+4x+3}{(x^2+1)(x^2+4)}dx\\ & =\int \frac{x+1}{x^2+1}+\frac{-1}{x^2+4}dx\\ & =\int \frac{x}{x^2+1}dx+\int \frac{1}{x^2+1}dx+\int \frac{-1}{x^2+4}dx\\ & =\frac{\ln (x^2+1)}{2}+{\tan }^{-1}x-\frac{1}{2}{\tan }^{-1}\frac{x}{2}\\ & \\ & 2.\int \frac{1}{x+x\sqrt{x}}dx\\ & =2\int \frac{1}{\sqrt{x}+x}d\sqrt{x}=2\int (\frac{A}{\sqrt{x}}+\frac{B}{\sqrt{x}+1})d\sqrt{x}\\ & A=1B=-1\\ & =2\int (\frac{1}{\sqrt{x}}+\frac{-1}{\sqrt{x}+1})d\sqrt{x}\\ & =2(\int \frac{1}{\sqrt{x}}d\sqrt{x}-\int \frac{1}{\sqrt{x}+1}d\sqrt{x})\\ & =2(\ln \sqrt{x}-\ln (\sqrt{x}+1))\\ & =2\ln \frac{\sqrt{x}}{\sqrt{x}+1}\end{array}$