计算 $\int\frac{x+3}{x^2-5x+6}dx$

设 $\frac{x+3}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}$

解得 $\frac{x+3}{x^2-5x+6}=\frac{-5}{x-2}+\frac{6}{x-3}$

$\int\frac{x+3}{x^2-5x+6}dx=-5\ln|x-2|+6\ln|x-3|+C$

计算 $\int\frac{1}{x(x-1)^2}dx$

$\frac{1}{x(x-1)^2}=\frac{1}{x}+\frac{1}{(x-1)^2}-\frac{1}{x-1}$

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

计算 $\int\frac{x-2}{x^2+2x+3}dx$

$\frac{x-2}{x^2+2x+3}=\frac{1}{2}\frac{2x+2}{x^2+2x+3}-3\frac{1}{x^2+2x+3}$

$\int\frac{2x+2}{x^2+2x+3}dx=\ln(x^2+2x+3)+C$

$\int \frac{1}{x^2+2x+3}dx=\int \frac{1}{(x+1)^2+(\sqrt 2)^2}dx=\frac{1}{\sqrt{2}}\arctan\frac{x+1}{\sqrt 2}+C$

计算以下不定积分

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

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

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

用万能代换计算 $\int\frac{1}{\sin^4x}dx$

$\int\frac{1}{\sin^4x}dx=\int\frac{(1+u^2)^4}{16u^4}\cdot\frac{2}{1+u^2}du=\int\frac{(1+u^2)^3}{8u^4}du$

$=\int\frac{1+3u^2+3u^4+u^6}{8u^4}du=\frac{1}{8}(-\frac{1}{3u^3}-\frac{3}{u}+3u+\frac{u^3}{3})+C$

$u=\tan\frac{x}{2}$代回上式

不用万能代换计算 $\int\frac{1}{\sin^4x}dx$

$\int\frac{1}{\sin^4x}dx=\int\csc^2x(1+\cot^2x)dx$

$\int\csc^2xdx=-\cot x+C$

$\int\csc^2x\cot^2xdx=-\frac{1}{3}\cot^3 x+C$

计算以下不定积分

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

$\int\frac{\cot x}{1+\sin x}dx$

$\int\frac{1}{(2+\cos x)\sin x}dx$

计算 $\int\frac{1}{1+\sqrt[3]{x+2}}dx$

令$t=\sqrt[3]{x+2}$，则$x=t^3-2$，从而$dx=3t^2dt$

$\int\frac{1}{1+\sqrt[3]{x+2}}dx=\int\frac{3t^2}{1+t}dt=3\int(t-1)+\frac{1}{t+1}dt$

$=3(\frac{t^2}{2}-t+\ln|1+t|)+C$，再将$t=\sqrt[3]{x+2}$代回

计算 $\int\frac{1}{(1+\sqrt[3]{x})\sqrt x}dx$

令$t=\sqrt[6]{x}$，则$x=t^6$，从而$dx=6t^5dt$

$\int\frac{1}{(1+\sqrt[3]{x})\sqrt x}dx=\int\frac{6t^5}{(1+t^2)t^3}dt=6\int\frac{t^2}{1+t^2}dt=6\int(1-\frac{1}{1+t^2})dt$

$=6t-\arctan t+C$，再将$t=\sqrt[6]{x}$代回

计算以下不定积分

$\int\frac{1}{\sqrt{1+x}+\sqrt[3]{1+x}}dx$

$\int\frac{1}{x}\sqrt{\frac{1+x}{x}}dx$

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