$$$\frac{1}{a^{2} + x^{2}}$$$ 對 $$$x$$$ 的積分

求$$$\int \frac{1}{a^{2} + x^{2}}\, dx$$$。

令 $$$u=\frac{x}{\left|{a}\right|}$$$。

則 $$$du=\left(\frac{x}{\left|{a}\right|}\right)^{\prime }dx = \frac{dx}{\left|{a}\right|}$$$ (步驟見»)，並可得 $$$dx = \left|{a}\right| du$$$。

所以，

$${\color{red}{\int{\frac{1}{a^{2} + x^{2}} d x}}} = {\color{red}{\int{\frac{\left|{a}\right|}{a^{2} \left(u^{2} + 1\right)} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{\left|{a}\right|}{a^{2}}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$：

$${\color{red}{\int{\frac{\left|{a}\right|}{a^{2} \left(u^{2} + 1\right)} d u}}} = {\color{red}{\frac{\left|{a}\right| \int{\frac{1}{u^{2} + 1} d u}}{a^{2}}}}$$

$$$\frac{1}{u^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$：

$$\frac{\left|{a}\right| {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{a^{2}} = \frac{\left|{a}\right| {\color{red}{\operatorname{atan}{\left(u \right)}}}}{a^{2}}$$

回顧一下 $$$u=\frac{x}{\left|{a}\right|}$$$：

$$\frac{\left|{a}\right| \operatorname{atan}{\left({\color{red}{u}} \right)}}{a^{2}} = \frac{\left|{a}\right| \operatorname{atan}{\left({\color{red}{\frac{x}{\left|{a}\right|}}} \right)}}{a^{2}}$$

因此，

$$\int{\frac{1}{a^{2} + x^{2}} d x} = \frac{\left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{2}}$$

加上積分常數：

$$\int{\frac{1}{a^{2} + x^{2}} d x} = \frac{\left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{2}}+C$$

$$$\int \frac{1}{a^{2} + x^{2}}\, dx = \frac{\left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{2}} + C$$$A