几种基本形式的积分

在计算积分时通常会出现以下几种形式的积分：

$$\begin{array}{ll}{I}_1=\int \frac{\mathrm{d}x}{a^2{x}^2+b^2}& I_2=\int \frac{\mathrm{d}x}{a^2{x}^2-b^2}\\ I_3=\int \frac{\mathrm{d}x}{\sqrt{a^2{x}^2\pm b^2}}& I_4=\int \frac{\mathrm{d}x}{\sqrt{b^2-a^2{x}^2}}\\ I_5=\int \sqrt{a^2{x}^2\pm b^2}\mathrm{d}x& I_6=\int \sqrt{b^2-a^2{x}^2}\mathrm{d}x\end{array}$$

本文将对这几种积分的结果进行推导。

逐个求解

先来看最简单的 $I_1$ 和 $I_2$：

$$\begin{array}{rl}{I}_1& =\int \frac{\mathrm{d}x}{a^2{x}^2+b^2}=\int \frac{\frac{b}{a}\mathrm{d}\left(\frac{ax}{b}\right)}{b^2\left(1+{\left(\frac{ax}{b}\right)}^2\right)}=\frac{1}{ab}\arctan \frac{ax}{b}+C\\ I_2& =\int \frac{\mathrm{d}x}{a^2{x}^2-b^2}=\int \frac{1}{2b}\left(\frac{1}{ax-b}-\frac{1}{ax+b}\right)\mathrm{d}x\\ & =\frac{1}{2b}\left[\frac{1}{a}\ln \left|ax-b\right|-\frac{1}{a}\ln \left|ax+b\right|\right]\\ & =\frac{1}{2ab}\ln \left|\frac{ax-b}{ax+b}\right|+C\end{array}$$

这两个积分均不需要换元就能积出。

对于 $I_3=\int \frac{\mathrm{d}x}{\sqrt{a^2{x}^2\pm b^2}}$，我们应该分两种情况考虑：

$$\begin{array}{rl}{I}_3^{+}& =\int \frac{\mathrm{d}x}{\sqrt{a^2{x}^2+b^2}}\stackrel{x=\frac{b}{a}\sinh t}{=}\int \frac{\frac{b}{a}\cosh t\mathrm{d}t}{b\cosh t}=\frac{t}{a}=\frac{1}{a}{\sinh }^{-1}\frac{ax}{b}\\ & =\frac{1}{a}\ln \left(\frac{ax}{b}+\sqrt{{\left(\frac{ax}{b}\right)}^2+1}\right)=\frac{1}{a}\ln \left(ax+\sqrt{a{x}^2+b^2}\right)-\frac{1}{a}\ln b\\ & =\frac{1}{a}\ln \left(ax+\sqrt{a{x}^2+b^2}\right)+C\\ I_3^{-}& =\int \frac{\mathrm{d}x}{\sqrt{a^2{x}^2-b^2}}\stackrel{x=\frac{b}{a}\cosh t}{=}\int \frac{\frac{b}{a}\sinh t\mathrm{d}t}{b\sinh t}=\frac{t}{a}=\frac{1}{a}{\cosh }^{-1}\frac{ax}{b}\\ & =\frac{1}{a}\ln \left(ax+\sqrt{a^2{x}^2-b^2}\right)+C\end{array}$$

于是

$$I_3=\frac{1}{a}\ln \left(ax+\sqrt{a^2{x}^2\pm b^2}\right)+C$$

反双曲函数

因为

$$\begin{array}{rl}y& =\sinh x=\frac{{\text{e}}^x-{\text{e}}^{-x}}{2}\\ \Rightarrow x& ={\sinh }^{-1}y\end{array}$$

满足

$${\text{e}}^{2x}-2y{\text{e}}^x-1=0$$

用求根公式（舍负值）解得

$${\text{e}}^x=\frac{2y\pm \sqrt{4y^2+4}}{2}=y+\sqrt{y^2+1}$$

取对数，$x=\ln \left(y+\sqrt{y^2+1}\right)$ ，即

$${\sinh }^{-1}y=\ln \left(y+\sqrt{y^2+1}\right)$$

同理有 ${\cosh }^{-1}y=\ln \left(y+\sqrt{y^2-1}\right)$

三角换元

如果不熟悉双曲函数也可以用三角函数：

$$\begin{array}{rl}{I}_3^{+}& =\int \frac{\mathrm{d}x}{\sqrt{a^2{x}^2+b^2}}\stackrel{x=\frac{b}{a}\tan t}{=}\int \frac{\frac{b}{a}{\sec }^{2}t\mathrm{d}t}{b\sec t}=\frac{1}{a}\int \sec t\mathrm{d}t\\ & =\frac{1}{a}\ln \left|\sec t+\tan t\right|=\frac{1}{a}\ln \left|\sqrt{1+{\left(\frac{ax}{b}\right)}^2}+\frac{ax}{b}\right|\\ & =\frac{1}{a}\ln \left(ax+\sqrt{a^2{x}^2+b^2}\right)+C\end{array}$$

$$\begin{array}{rl}{I}_3^{-}& =\int \frac{\mathrm{d}x}{\sqrt{a^2{x}^2-b^2}}\stackrel{x=\frac{b}{a}\sec t}{=}\int \frac{\frac{b}{a}\sec t\tan t\mathrm{d}t}{b\tan t}=\frac{1}{a}\int \sec t\mathrm{d}t\\ & =\frac{1}{a}\ln \left|\sec t+\tan t\right|=\frac{1}{a}\ln \left|\frac{ax}{b}+\sqrt{{\left(\frac{ax}{b}\right)}^2-1}\right|\\ & =\frac{1}{a}\ln \left(ax+\sqrt{a^2{x}^2-b^2}\right)+C\end{array}$$

$I_4$ 也比较简单，凑微分可得

$$I_4=\int \frac{\mathrm{d}x}{\sqrt{b^2-a^2{x}^2}}=\int \frac{\frac{b}{a}\mathrm{d}\left(\frac{ax}{b}\right)}{b\sqrt{1-{\left(\frac{ax}{b}\right)}^2}}=\frac{1}{a}\arcsin \frac{ax}{b}+C$$

最后两个，用分部积分法可化为之前求过的积分：

$$\begin{array}{rl}{I}_5& =\int \sqrt{a^2{x}^2\pm b^2}\mathrm{d}x=x\sqrt{a^2{x}^2\pm b^2}-\int \frac{2a^2{x}^2}{2\sqrt{a^2{x}^2\pm b^2}}\mathrm{d}x\\ & =x\sqrt{a^2{x}^2\pm b^2}-\int \frac{a^2{x}^2\pm b^2\mp b^2}{\sqrt{a^2{x}^2\pm b^2}}\mathrm{d}x\\ & =x\sqrt{a^2{x}^2\pm b^2}-I_5\pm b^2{I}_3\\ & =\frac{x}{2}\sqrt{a^2{x}^2\pm b^2}\pm \frac{b^2}{2a}\ln \left(ax+\sqrt{a^2{x}^2\pm b^2}\right)+C\end{array}$$

$$\begin{array}{rl}{I}_6& =\int \sqrt{b^2-a^2{x}^2}\mathrm{d}x=x\sqrt{b^2-a^2{x}^2}-\int \frac{-2a^2{x}^2}{2\sqrt{b^2-a^2{x}^2}}\mathrm{d}x\\ & =x\sqrt{b^2-a^2{x}^2}-\int \frac{\left(b^2-a^2{x}^2\right)-b^2}{\sqrt{b^2-a^2{x}^2}}\mathrm{d}x\\ & =x\sqrt{b^2-a^2{x}^2}-I_6+b^2{I}_4\\ & =\frac{x}{2}\sqrt{b^2-a^2{x}^2}+\frac{b^2}{2a}\arcsin \frac{ax}{b}+C\end{array}$$

总结

以下 8 个积分（6 个公式）可直接使用：

$$\begin{array}{rl}\int \frac{\mathrm{d}x}{a^2{x}^2+b^2}& =\frac{1}{ab}\arctan \frac{ax}{b}+C\\ \int \frac{\mathrm{d}x}{a^2{x}^2-b^2}& =\frac{1}{2ab}\ln \left|\frac{ax-b}{ax+b}\right|+C\end{array}$$

$$\begin{array}{rl}\int \frac{\mathrm{d}x}{\sqrt{a^2{x}^2\pm b^2}}& =\frac{1}{a}\ln \left(ax+\sqrt{a^2{x}^2\pm b^2}\right)+C\\ \int \frac{\mathrm{d}x}{\sqrt{b^2-a^2{x}^2}}& =\frac{1}{a}\arcsin \frac{ax}{b}+C\end{array}$$

$$\begin{array}{rl}\int \sqrt{a^2{x}^2\pm b^2}\mathrm{d}x& =\frac{x}{2}\sqrt{a^2{x}^2\pm b^2}\pm \frac{b^2}{2a}\ln \left(ax+\sqrt{a^2{x}^2\pm b^2}\right)+C\\ \int \sqrt{b^2-a^2{x}^2}\mathrm{d}x& =\frac{x}{2}\sqrt{b^2-a^2{x}^2}+\frac{b^2}{2a}\arcsin \frac{ax}{b}+C\end{array}$$