# 可化为有理函数的积分

1，无理函数（根式函数）：先变量代换，再利用有理函数的积分法

$$\displaystyle\int \frac{\sqrt{x+4}}{x}dx=\int \frac{u}{u^2-4}\cdot 2udu=\int \frac{2u^2}{u^2-4}du$$

$$\displaystyle\qquad\qquad\qquad= 2\int \frac{u^2-4+4}{u^2-4}du=2\int \left(1+\frac{4}{u^2-4}\right)du$$

$$\displaystyle\qquad\qquad\qquad= 2u+8\int \frac{du}{u^2-4}$$

$$\displaystyle\qquad\qquad\qquad\quad=2\ln|u-2|-2\ln|u+2|+C$$

$$\displaystyle\qquad\qquad\qquad\quad=2\ln |\frac{u-2}{u+2}|+C$$

$$\displaystyle\qquad\qquad\qquad=2\sqrt {x+4}+2\ln|\frac{\sqrt{x+4}-2}{\sqrt {x+4}+2}|+C$$

$$\displaystyle\quad\int \frac{dx}{(1+\sqrt[3]{x})\sqrt{x}}=\int \frac{6u^2du}{(1+u)^2 u^3}=\int \frac{6u^2}{1+u^2}du$$

$$\displaystyle\qquad\qquad\qquad\qquad=6\int \frac{u^2+1-1}{u^2+1}du=6\int \left(1-\frac{1}{u^2+1}\right)du$$

$$\displaystyle\qquad\qquad\qquad\qquad=6u-6\int \frac{du}{1+u^2}=6u-6\arctan u+C$$

$$\displaystyle\qquad\qquad\qquad\qquad=6\sqrt[6]{u}-6\arctan \sqrt[6]{u}+C$$

$$\displaystyle\quad\Rightarrow\quad \frac{1}{x}=u^2-1\quad\Rightarrow\quad x=\frac{1}{u^2-1}\quad\Rightarrow\quad dx=\frac{2udu}{\left(u^2-1\right)^2}$$

$$\displaystyle\int \frac{1}{x}\sqrt {\frac{x+1}{x}}dx=\int (u^2-1)\cdot u\cdot\frac{2u}{\left(u^2-1\right)^2}du$$

$$\displaystyle\qquad\qquad\qquad\quad=\int \frac{2u^2}{u^2-1}du$$

$$\displaystyle\qquad\qquad\qquad\quad=2\int \frac{u^2-1+1}{u^2-1}du$$

$$\displaystyle\qquad\qquad\qquad\quad=2\int \left(1+\frac{1}{u^2-1}\right)du$$

$$\displaystyle\qquad\qquad\qquad\quad=2u+2\int \frac{du}{u^2-1}$$

$$\displaystyle\qquad\qquad\qquad\quad=\ln|u-1|-\ln|u+1|+C=\ln|\frac{u-1}{u+1}|+C$$

$$\displaystyle\qquad\qquad\qquad=2\sqrt {\frac{x+1}{x}}+ln|\frac{\sqrt{\frac{x+1}{x}}-1}{\sqrt{\frac{x+1}{x}}+1}|+C$$

2，三角函数

$$\displaystyle\quad\sin x=2 \sin\frac{x}{2}\cos \frac{x}{2}=2\tan \frac{x}{2}\cdot\cos^2 \frac{x}{2}$$

$$\displaystyle\qquad\quad\ =\frac{2\tan \frac{x}{2}}{\sec^2 \frac{x}{2}}=\frac{2\tan \frac{x}{2}}{\tan^2 \frac{x}{2}+1}=\frac {2t}{1+t^2}$$

$$\displaystyle\quad\cos x=\cos^2 \frac{x}{2}-\sin^2\frac{x}{2}$$

$$\displaystyle\qquad\quad\ =\frac{1}{\sec^2 \frac{x}{2}}-\frac{\sin^2\frac{x}{2}}{\cos^2 \frac{x}{2}}\cdot \cos^2\frac{x}{2}$$

$$\displaystyle\qquad\quad\ =\frac{1}{\sec^2 \frac{x}{2}}-\frac{\tan^2 \frac{x}{2}}{\sec^2\frac{x}{2}}$$

$$\displaystyle\qquad\quad\ =\frac{1-\tan^2\frac{x}{2}}{\sec^2 \frac{x}{2}}=\frac{1-t^2}{1+t^2}$$

$$\displaystyle\quad\int \frac{1+\sin x}{\sin x(1+\cos x)}dx=\int \frac{1+\frac{2t}{1+t^2}}{\frac{2t}{1+t^2}\left(1+\frac{1-t^2}{1+t^2}\right)}\cdot \frac{2dt}{1+t^2}$$

$$\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\int\frac{2(1+t^2+2t)}{2t(1+t^2+1-t^2)}dt$$

$$\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\int \frac{t^2+2t+1}{2t}dt$$

$$\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\frac{1}{2}\int (t+2+\frac{1}{t})dt$$

$$\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\frac{1}{2}(\frac{t^2}{2}+2t+\ln|t|)+C$$

$$\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\frac{t^2}{4}+t+\frac{1}{2}\ln|t|+C$$

$$\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\frac{\tan^2 \frac{x}{2}}{4}+\tan \frac{x}{2}+\frac{1}{2}\ln|\tan\frac{x}{2}|+C$$