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