双元法出自知乎答主 “虚调子”，本质上是一种特殊的凑微分手段，用来求解分式带根号的不定积分十分方便，在我下一篇有关莫比乌斯变换的笔记中要经常用到，所以先写一篇关于双元法的笔记吧

实圆微分式

对于实圆：
$$
x^2+y^2=A^2
$$
取微分：
$$
\begin{align*}
x, \mathrm{d}x&+y, \mathrm{d}y=0\
\frac{, \mathrm{d}x}{y}&=-\frac{, \mathrm{d}y}{x}\
\frac{, \mathrm{d}x}{iy}&=\frac{, \mathrm{d}(iy)}{x}=\frac{, \mathrm{d}(x+iy)}{x+iy}\
\frac{, \mathrm{d}x}{y}&=\frac{y, \mathrm{d}x-x, \mathrm{d}y}{y^2+x^2}
\end{align*}
$$

虚圆微分式

对于虚圆：
$$
x^2-y^2=A^2 \quad or \quad y^2-x^2=A^2
$$
取微分：
$$
\begin{align*}
x, \mathrm{d}x&-y, \mathrm{d}y=0\
\frac{, \mathrm{d}x}{y}&=\frac{, \mathrm{d}y}{x}=\frac{, \mathrm{d}x+, \mathrm{d}y}{x+y}\
\frac{, \mathrm{d}x}{y}&=\frac{y, \mathrm{d}x-x, \mathrm{d}y}{y^2-x^2}
\end{align*}
$$

双元第一公式：

Re:

$$
\int \frac{, \mathrm{d}x}{y} = \int \frac{y , \mathrm{d}x}{y^2}
$$

合分比由
$$
\frac{y , \mathrm{d}x}{y^2} = -\frac{x , \mathrm{d}y}{x^2} \left( \frac{, \mathrm{d}x}{y} = -\frac{, \mathrm{d}y}{x} \right)
$$
有
$$
= \int \frac{y , \mathrm{d}x - x , \mathrm{d}y}{x^2 + y^2} = \int \frac{y^2}{x^2 + y^2} , \mathrm{d} \left( \frac{x}{y} \right) = \int \frac{1}{1 + \left( \frac{x}{y} \right)^2} , \mathrm{d} \left( \frac{x}{y} \right) = \arctan \frac{x}{y}
$$

Im:

$$
\int \frac{, \mathrm{d}x}{y} = \int \frac{, \mathrm{d}x + , \mathrm{d}y}{y + x} = \int \frac{, \mathrm{d}(x + y)}{x + y} = \ln (x + y)
$$

故
$$
\int \frac{, \mathrm{d}x}{y} =
\begin{cases}
\arctan \displaystyle \frac{x}{y}, & \text{Re} \
\ln (x + y), & \text{Im}
\end{cases}
$$

双元第三公式

Re：

$$
\begin{align*}
\int \frac{, \mathrm{d}x}{y^3}&=\int \frac{1}{y^2} \cdot \frac{, \mathrm{d}x}{y}=\int \frac{1}{y^2} \cdot \frac{y, \mathrm{d}x}{y^2}=\int \frac{1}{y^2} \cdot \frac{y, \mathrm{d}x - x, \mathrm{d}y}{y^2 + x^2}\
&=\frac{1}{y^2 + x^2} \int \frac{y, \mathrm{d}x - x, \mathrm{d}y}{y^2}=\frac{1}{y^2 + x^2} \int , \mathrm{d}\left( \frac{x}{y} \right) \
&=\frac{1}{y^2 + x^2} \cdot \frac{x}{y}
\end{align*}
$$

Im:

$$
\begin{align*}
\int \frac{, \mathrm{d}x}{y^3}&=\int \frac{1}{y^2} \cdot \frac{, \mathrm{d}x}{y}=\int \frac{1}{y^2} \cdot \frac{y, \mathrm{d}x}{y^2}=\int \frac{1}{y^2} \cdot \frac{y, \mathrm{d}x - x, \mathrm{d}y}{y^2 - x^2}\
&=\frac{1}{y^2 - x^2} \int \frac{y, \mathrm{d}x - x, \mathrm{d}y}{y^2}=\frac{1}{y^2 - x^2} \int , \mathrm{d}\left( \frac{x}{y} \right)\
&=\frac{1}{y^2 - x^2} \cdot \frac{x}{y}
\end{align*}
$$

故
$$
\int \frac{, \mathrm{d}x}{y^3} =
\begin{cases}
\displaystyle \frac{1}{y^2 + x^2} \cdot \frac{x}{y}, & \text{Re} \
\displaystyle \frac{1}{y^2 - x^2} \cdot \frac{x}{y}, & \text{Im}
\end{cases}
$$
总括为
$$
\int \frac{, \mathrm{d}x}{y^3} = \frac{1}{y^2 \pm x^2} \cdot \frac{x}{y}
$$

双元乘积公式

$$
\begin{align}
\int y, \mathrm{d}x &=\frac{\displaystyle \int (y, \mathrm{d}x + x, \mathrm{d}y) + \int (y, \mathrm{d}x - x, \mathrm{d}y)}{2}\
&= \frac{\displaystyle \int , \mathrm{d}(xy)}{2} + \frac{\displaystyle \int (y, \mathrm{d}x - x, \mathrm{d}y)}{2}\
&=\frac{1}{2}xy + \frac{\displaystyle \int (y, \mathrm{d}x - x, \mathrm{d}y)}{2}\
\end{align}
$$

其中
$$
\begin{align}
\frac{1}{2} \int (y, \mathrm{d}x - x, \mathrm{d}y) &= \frac{y^2 \pm x^2}{2} \int \frac{y, \mathrm{d}x - x, \mathrm{d}y}{y^2 \pm x^2} \
&= \frac{y^2 \pm x^2}{2} \int \frac{, \mathrm{d}x}{y}
\end{align}
$$
最终形式
$$
\int y, \mathrm{d}x = \frac{1}{2}xy + \frac{y^2 \pm x^2}{2} \int \frac{, \mathrm{d}x}{y}
$$

双元点火不足公式

对于
$$
y^2 \pm x^2=A^2
$$
递推公式：
$$
\int y^n , \mathrm{d}x = \frac{1}{n + 1}(y^n x) + \frac{n}{n + 1}(A) \int y^{n - 2} , \mathrm{d}x\
\int \frac{, \mathrm{d}x}{y^n} = \frac{1}{n - 2}\cdot\frac{x}{y^{n - 2}A} + \frac{n - 3}{n - 2}\cdot\frac{1}{A} \int \frac{, \mathrm{d}x}{y^{n - 2}}
$$
推论（特殊情况）：
$$
\int y , \mathrm{d}x = \frac{1}{2}(xy) + \frac{1}{2}(A) \int \frac{, \mathrm{d}x}{y}\
\int \frac{, \mathrm{d}x}{y^3} = \frac{1}{A}\cdot\frac{x}{y}
$$

其余常见构型

$$
\begin{align*}
\int \frac{, \mathrm{d}x}{x^2y}&=-\frac{1}{A}\cdot \frac{y}{x}\
\int \frac{, \mathrm{d}x}{xy^2}&=-\frac{1}{A}\cdot \ln\frac{y}{x}=\frac{1}{A}\cdot \ln\frac{x}{y}
\end{align*}
$$

典型例题

我们先从简单的基本积分开始，熟悉双元法的解题流程

普通双元配凑

一、

$$
\int \frac{, \mathrm{d}x}{\sqrt{x^2+a^2}}
$$

令 $y=\sqrt{x^2+a^2}$，则有 $y^2-x^2=a^2$，取微分得 $x, \mathrm{d}x=y, \mathrm{d}y$

且有
$$
\frac{, \mathrm{d}x}{y}=\frac{, \mathrm{d}y}{x}=\frac{, \mathrm{d}\left( x+y \right)}{x+y}
$$
则原式为
$$
\begin{align}
\int \frac{, \mathrm{d}x}{y}=\frac{, \mathrm{d}\left( x+y \right)}{x+y} &=\ln |x+y|+C\
&=\ln|x+\sqrt{x^2+a^2}|+C
\end{align}
$$

二、

$$
\int \frac{, \mathrm{d}x}{x\sqrt{a^2-x^2}}
$$

令 $y=\sqrt{a^2-x^2}$

则原式为
$$
\begin{align}
\int \frac{, \mathrm{d}x}{x\sqrt{a^2-x^2}} &=\int \frac{, \mathrm{d}x}{xy}=-\int \frac{, \mathrm{d}y}{x^2}\
&=\int \frac{, \mathrm{d}y}{y^2-a^2}=\frac{1}{2a}\ln|\frac{\sqrt{a^2-x^2}-a}{\sqrt{a^2-x^2}+a}|+C
\end{align}
$$

三、

$$
\int \ln\left( \sqrt{x+1}-\sqrt{1-x} \right)dx
$$

令 $m=\sqrt{x+1},m=\sqrt{1-x}$，则有 $m^2+n^2=2$，取微分得 $x, \mathrm{d}x=y, \mathrm{d}y$

则原式为
$$
\begin{align}
\int \ln\left( \sqrt{x+1}+\sqrt{1-x} \right), \mathrm{d}x &=x\ln(m+n)-\int x, \mathrm{d}\ln(m+n)\
&=x\ln(m+n)-x\cdot \frac{, \mathrm{d}m+, \mathrm{d}n}{m+n}\
&=x\ln(m+n)-\frac{1}{2}\int (m-n)(, \mathrm{d}m+, \mathrm{d}n)\
&=x\ln(m+n)-\frac{1}{4}(m^2+n^2)+\int \frac{n, \mathrm{d}m-m, \mathrm{d}n}{2}\
&=x\ln (m+n)-\frac{1}{2}x+\int \frac{n^2, \mathrm{d}\left( \frac{m}{n} \right)}{m^2+n^2}\
&=x\ln (m+n)-\frac{1}{2}x+\int \frac{, \mathrm{d}\left( \frac{m}{n} \right)}{\left( \frac{m}{n} \right)^2+1}\
&=x\ln\left( \sqrt{x+1}+\sqrt{1-x} \right)-\frac{1}{2}x+\arctan{\sqrt{\frac{1+x}{1-x}}}+C
\end{align}
$$

四、

$$
\int \frac{, \mathrm{d}x}{\sqrt{x^2+2x+3}}
$$

令
$$
m=x+1,\quad n=\sqrt{x^2+2x+3}
$$
原式为
$$
\begin{align}
\int \frac{, \mathrm{d}m}{n}&=\int \frac{, \mathrm{d}\left(m+n\right)}{m+n}\
&=\ln|m+n|+C\
&=\ln|x+1+\sqrt{x^2+2x+3}|+C
\end{align}
$$

五、

$$
\int \sqrt{x^2+2x+3}
$$

同上题所示设双元

原式为
$$
\begin{align}
\int ndm&=\frac{1}{2}mn+\frac{n^2-m^2}{2}\int \frac{, \mathrm{d}m}{n}\
&=\frac{1}{2}\left(x+1\right)\sqrt{x^2+2x+3}+\ln|x+1+\sqrt{x^2+2x+3}|+C
\end{align}
$$

六、

$$
\int \frac{x, \mathrm{d}x}{\sqrt{-x^2-2x+3}}
$$

令
$$
m=x+1,\quad n=\sqrt{-x^2-2x+3}
$$
原式为
$$
\begin{align}
\int \frac{\left(m-1\right), \mathrm{d}m}{n} &= \int \frac{m, \mathrm{d}m}{n}-\frac{, \mathrm{d}m}{n}\
&=-\int , \mathrm{d}n-\arcsin \frac{m}{2}\
&=-\sqrt{-x^2-2x+3}-\arcsin\frac{x+1}{2}+C
\end{align}
$$
接下来是一些技巧性配凑

对勾双元配凑

七、

$$
\int \frac{, \mathrm{d}x}{1+x^4}
$$

令
$$
m=x+\displaystyle \frac{1}{x},\quad n=x-\frac{1}{x}
$$
原式为
$$
\begin{align}
\int \frac{, \mathrm{d}x}{1+x^4} &=\int \frac{\frac{, \mathrm{d}x}{x^2}}{x^2+\frac{1}{x^2}}=\int \frac{, \mathrm{d}\left(-\frac{1}{x}\right)}{x^2+\frac{1}{x^2}}\
&=\int \frac{, \mathrm{d}\left(\frac{n-m}{2}\right)}{m^2-2}=\frac{1}{2}\int \frac{, \mathrm{d}n}{n^2+2}-\frac{1}{2}\int \frac{, \mathrm{d}m}{m^2-2}\
&=\frac{1}{2\sqrt{2}}\arctan\frac{x-\frac{1}{x}}{\sqrt{2}}-\frac{1}{4\sqrt{2}}\ln|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}|+C
\end{align}
$$
注：为确保严谨，此处最好用如下公式进行间断点修正
$$
\begin{align*}
\arctan\left(k\left(x-\frac{1}{x}\right)\right) &= \arctan\left(k\left(x-\frac{1}{x}\right)\right)

\arctan\left(\frac{1}{kx}\right) + \arctan\left(kx\right) \
&= \arctan\left(kx^3 + \left(\frac{1}{k}-1\right)x\right) + \arctan\left(kx\right)
\end{align*}
$$

八、

对勾三元
$$
\int \frac{1 - x^2}{1 + x^2} \frac{, \mathrm{d}x}{\sqrt{x^4 + x^2 + 1}}
$$

令

$$
p = x + \frac{1}{x}, \quad q = x - \frac{1}{x}, \quad r = \frac{\sqrt{x^4 + x^2 + 1}}{x}
$$

原式为
$$
\begin{align}
\int \frac{1 - x^2}{1 + x^2} \frac{, \mathrm{d}x}{\sqrt{x^4 + x^2 + 1}} &= - \int \frac{, \mathrm{d}p}{pr}\
&= - \int \frac{, \mathrm{d}r}{p^2} \
&= - \int \frac{, \mathrm{d}r}{r^2 + 1}\
&= \arctan \left( \frac{x}{\sqrt{x^4 + x^2 + 1}} \right) + C
\end{align}
$$

三角三元

九、

$$
\int \frac{\sin^2 x \cos x}{\sin x + \cos x} , \mathrm{d}x
$$

令
$$
p = \sin x + \cos x, \quad q = \sin x - \cos x, \quad r = \sqrt{2}
$$

原式为
$$
\begin{align}
\int \frac{\sin^2 x \cos x}{\sin x + \cos x} , \mathrm{d}x &= \int \frac{\frac{r^2}{2} \cdot \frac{p-q}{2} , , \mathrm{d}p}{pq}\
&= \frac{1}{4} \int \frac{r^2 , \mathrm{d}p}{q} - \frac{1}{4} \int \frac{r^2 , \mathrm{d}p}{p}\
&= \frac{1}{4} \int \frac{, \mathrm{d}p}{q} - \frac{1}{4} \int q , \mathrm{d}p - \frac{1}{4} \int \frac{p^2 - 1}{p} , \mathrm{d}p\
&= -\frac{1}{8} \int p , \mathrm{d}q + q , \mathrm{d}p - \frac{1}{4} \left( \frac{1}{2} p^2 - \ln p \right)\
&= \frac{1}{4} \ln p - \frac{1}{8} \left( pq + p^2 \right)\
&= \frac{1}{4} \ln \left( \sin x + \cos x \right) - \frac{1}{4} \cos x \left( \sin x + \cos x \right) + C
\end{align}
$$

其他双元配凑

十、

$$
\int \frac{, \mathrm{d}x}{x - \sqrt{x^2 + 2x}}
$$

令
$$
p = x + 1 + \sqrt{x^2 + 2x}, \quad q = x + 1 - \sqrt{x^2 + 2x}
$$

原式为
$$
\begin{align}
\int \frac{, \mathrm{d}x}{x - \sqrt{x^2 + 2x}} &= \int \frac{1}{q-1} , \mathrm{d}\left( \frac{p+q}{2} \right)\
&= \frac{1}{2} \int \frac{, \mathrm{d}p}{q-1} + \frac{1}{2} \int \frac{, \mathrm{d}q}{q-1}\
&= \frac{1}{2} \int \frac{p , \mathrm{d}p}{1-p} + \frac{1}{2} \ln |q-1|\
&= -\frac{1}{2} \int \frac{p-1+1}{p-1} , \mathrm{d}p + \frac{1}{2} \ln |q-1|\
&= -\frac{1}{2} \int , \mathrm{d}p - \frac{1}{2} \int \frac{, \mathrm{d}p}{p-1} + \frac{1}{2} \ln |q-1|\
&= -\frac{p}{2} - \frac{1}{2} \ln |p-1| + \frac{1}{2} \ln |q-1| + C\
&= \frac{1}{2} \ln \left| \frac{x - \sqrt{x^2 + 2x}}{x + \sqrt{x^2 + 2x}} \right| - \frac{x + 1 + \sqrt{x^2 + 2x}}{2} + C
\end{align}
$$

十一、

$$
\int \sqrt{\frac{e^x - 1}{e^x + 1}} , dx
$$

令
$$
m = e^x, \quad n = \sqrt{e^{2x} - 1}
$$

$$
dm = e^x, \mathrm{d}x = m , \mathrm{d}x \implies \mathrm{d}x = \frac{dm}{m}
$$

原式为
$$
\begin{align}
\int \sqrt{\frac{e^x - 1}{e^x + 1}} , \mathrm{d}x &= \int \frac{\sqrt{m - 1}}{\sqrt{m + 1}} \cdot \frac{, \mathrm{d}m}{m}\
&= \int \frac{m - 1}{n} \cdot \frac{, \mathrm{d}m}{m} \
&= \int \frac{, \mathrm{d}m}{n} - \int \frac{, \mathrm{d}m}{m n}\
&= \ln(m + n)-\int \frac{, \mathrm{d}n}{n^2 + 1} \
&= \ln(m + n)-\arctan n+C\
&= \ln\left(e^x + \sqrt{e^{2x} - 1}\right) - \arctan\left(\sqrt{e^{2x} - 1}\right) + C
\end{align}
$$

双元法还有许多延申（如下图所示），本文所介绍的方法足以解决绝大部分的简单不定积分，想继续深入了解的请关注"虚调子"大佬的知乎账号