有一些形式的被积函数，可以通过一些特殊的换元法，化成有理函数的积分。而我们知道，有理函数总是可以求出它的积分的。
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
例1，求\(\displaystyle\quad\int \frac{\sqrt{x+4}}{x}dx\)
解：令\(\displaystyle u=\sqrt{x+4}\)，\(\displaystyle x=u^2-4\)，\(\displaystyle dx=2udu\)
\(\displaystyle\int \frac{\sqrt{x+4}}{x}dx=\int \frac{u}{u^2-4}\cdot 2udu=\int \frac{2u^2}{u^2-4}du\)
\(\displaystyle\qquad\qquad\qquad= 2\int \frac{u^2-4+4}{u^2-4}du=2\int \left(1+\frac{4}{u^2-4}\right)du\)
\(\displaystyle\qquad\qquad\qquad= 2u+8\int \frac{du}{u^2-4}\)
由 \(\displaystyle\quad \frac{1}{u^2-4}=\frac{A}{u-2}+\frac{B}{u+2}\)
得\(\displaystyle\quad A(u+2)+B(u-2)=1\)
当\(\displaystyle\quad u=2\quad\ \ \ \Rightarrow\ \ \quad 4A=1\quad\ \Rightarrow\quad A=\frac{1}{4}\)
当\(\displaystyle\quad u=-2\quad\Rightarrow\quad -4B=1\quad\Rightarrow\quad B=-\frac{1}{4}\)
而 \(\displaystyle\quad 8\int \frac{du}{u^2-4}=2\int \frac{1}{u-2}du-2\int \frac{du}{u+2}\)
\(\displaystyle\qquad\qquad\qquad\quad=2\ln|u-2|-2\ln|u+2|+C\)
\(\displaystyle\qquad\qquad\qquad\quad=2\ln |\frac{u-2}{u+2}|+C\)
所以，原积分\(\displaystyle\ =2u+2\ln |\frac{u-2}{u+2}|+C\)
\(\displaystyle\qquad\qquad\qquad=2\sqrt {x+4}+2\ln|\frac{\sqrt{x+4}-2}{\sqrt {x+4}+2}|+C\)
例2，求\(\displaystyle\quad\int \frac{dx}{(1+\sqrt[3]{x})\sqrt{x}}\)
解：令\(\displaystyle\ u=\sqrt[6]{x}\)，\(\displaystyle x=u^6\)，\(\displaystyle dx=6u^5du\)
\(\displaystyle\quad\int \frac{dx}{(1+\sqrt[3]{x})\sqrt{x}}=\int \frac{6u^2du}{(1+u)^2 u^3}=\int \frac{6u^2}{1+u^2}du\)
\(\displaystyle\qquad\qquad\qquad\qquad=6\int \frac{u^2+1-1}{u^2+1}du=6\int \left(1-\frac{1}{u^2+1}\right)du\)
\(\displaystyle\qquad\qquad\qquad\qquad=6u-6\int \frac{du}{1+u^2}=6u-6\arctan u+C\)
\(\displaystyle\qquad\qquad\qquad\qquad=6\sqrt[6]{u}-6\arctan \sqrt[6]{u}+C\)
例3，求\(\displaystyle\quad\int \frac{1}{x}\sqrt {\frac{x+1}{x}}dx\)
解：令\(\displaystyle\quad u=\sqrt{\frac{x+1}{x}}\)，则\(\displaystyle\quad u^2=\frac{x+1}{x}=1+\frac{1}{x}\)
\(\displaystyle\quad\Rightarrow\quad \frac{1}{x}=u^2-1\quad\Rightarrow\quad x=\frac{1}{u^2-1}\quad\Rightarrow\quad dx=\frac{2udu}{\left(u^2-1\right)^2}\)
\(\displaystyle\int \frac{1}{x}\sqrt {\frac{x+1}{x}}dx=\int (u^2-1)\cdot u\cdot\frac{2u}{\left(u^2-1\right)^2}du\)
\(\displaystyle\qquad\qquad\qquad\quad=\int \frac{2u^2}{u^2-1}du\)
\(\displaystyle\qquad\qquad\qquad\quad=2\int \frac{u^2-1+1}{u^2-1}du\)
\(\displaystyle\qquad\qquad\qquad\quad=2\int \left(1+\frac{1}{u^2-1}\right)du\)
\(\displaystyle\qquad\qquad\qquad\quad=2u+2\int \frac{du}{u^2-1}\)
由\(\displaystyle\quad\frac{1}{u^2-1}=\frac{A}{u-1}+\frac{B}{u+1}\quad\Rightarrow\quad A(u+1)+B(u-1)=1\)
当\(\displaystyle\ u=1\quad\ \ \ \Rightarrow\quad\ A=\frac{1}{2}\)
当\(\displaystyle\ u=-1\quad\Rightarrow\quad B=-\frac{1}{2}\)
而\(\displaystyle\quad 2\int \frac{du}{u^2-1}=\int \left(\frac{1}{u-1}-\frac{1}{u+1}\right)du\)
\(\displaystyle\qquad\qquad\qquad\quad=\ln|u-1|-\ln|u+1|+C=\ln|\frac{u-1}{u+1}|+C\)
所以，原积分\(\displaystyle\ \ =2u+\ln|\frac{u-1}{u+1}|+C\)
\(\displaystyle\qquad\qquad\qquad=2\sqrt {\frac{x+1}{x}}+ln|\frac{\sqrt{\frac{x+1}{x}}-1}{\sqrt{\frac{x+1}{x}}+1}|+C\)
注：\(\displaystyle\ \sqrt{\frac{ax+b}{cx+d}}\ \)都可以直接换元。
2，三角函数
半角代换（万能代换）：\(\displaystyle\quad\tan \frac{x}{2}=t\)
\(\displaystyle\quad\sin x=2 \sin\frac{x}{2}\cos \frac{x}{2}=2\tan \frac{x}{2}\cdot\cos^2 \frac{x}{2}\)
\(\displaystyle\qquad\quad\ =\frac{2\tan \frac{x}{2}}{\sec^2 \frac{x}{2}}=\frac{2\tan \frac{x}{2}}{\tan^2 \frac{x}{2}+1}=\frac {2t}{1+t^2}\)
\(\displaystyle\quad\cos x=\cos^2 \frac{x}{2}-\sin^2\frac{x}{2}\)
\(\displaystyle\qquad\quad\ =\frac{1}{\sec^2 \frac{x}{2}}-\frac{\sin^2\frac{x}{2}}{\cos^2 \frac{x}{2}}\cdot \cos^2\frac{x}{2}\)
\(\displaystyle\qquad\quad\ =\frac{1}{\sec^2 \frac{x}{2}}-\frac{\tan^2 \frac{x}{2}}{\sec^2\frac{x}{2}}\)
\(\displaystyle\qquad\quad\ =\frac{1-\tan^2\frac{x}{2}}{\sec^2 \frac{x}{2}}=\frac{1-t^2}{1+t^2}\)
例4，求\(\displaystyle\quad\int \frac{1+\sin x}{\sin x(1+\cos x)}dx\)
解：令\(\displaystyle\quad t=\tan \frac{x}{2}\)，则\(\displaystyle\quad x=2\arctan t\)，\(\displaystyle\quad dx=\frac{2dt}{1+t^2}\)
\(\displaystyle\quad\int \frac{1+\sin x}{\sin x(1+\cos x)}dx=\int \frac{1+\frac{2t}{1+t^2}}{\frac{2t}{1+t^2}\left(1+\frac{1-t^2}{1+t^2}\right)}\cdot \frac{2dt}{1+t^2}\)
\(\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\int\frac{2(1+t^2+2t)}{2t(1+t^2+1-t^2)}dt\)
\(\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\int \frac{t^2+2t+1}{2t}dt\)
\(\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\frac{1}{2}\int (t+2+\frac{1}{t})dt\)
\(\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\frac{1}{2}(\frac{t^2}{2}+2t+\ln|t|)+C\)
\(\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\frac{t^2}{4}+t+\frac{1}{2}\ln|t|+C\)
\(\displaystyle\qquad\qquad\qquad\qquad\qquad\ =\frac{\tan^2 \frac{x}{2}}{4}+\tan \frac{x}{2}+\frac{1}{2}\ln|\tan\frac{x}{2}|+C\)