$$$\frac{1}{x^{2} \left(a^{2} + x^{2}\right)}$$$ 對 $$$x$$$ 的積分

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

進行部分分式分解:

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

逐項積分：

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

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

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

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

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

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

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

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 $$$dx = \left|{a}\right| du$$$。

該積分可改寫為

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

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 $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$：

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

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

$$- \frac{1}{a^{2} x} - \frac{\left|{a}\right| {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{a^{4}} = - \frac{1}{a^{2} x} - \frac{\left|{a}\right| {\color{red}{\operatorname{atan}{\left(u \right)}}}}{a^{4}}$$

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

$$- \frac{1}{a^{2} x} - \frac{\left|{a}\right| \operatorname{atan}{\left({\color{red}{u}} \right)}}{a^{4}} = - \frac{1}{a^{2} x} - \frac{\left|{a}\right| \operatorname{atan}{\left({\color{red}{\frac{x}{\left|{a}\right|}}} \right)}}{a^{4}}$$

因此，

$$\int{\frac{1}{x^{2} \left(a^{2} + x^{2}\right)} d x} = - \frac{1}{a^{2} x} - \frac{\left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{4}}$$

化簡：

$$\int{\frac{1}{x^{2} \left(a^{2} + x^{2}\right)} d x} = \frac{- a^{2} - x \left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{4} x}$$

加上積分常數：

$$\int{\frac{1}{x^{2} \left(a^{2} + x^{2}\right)} d x} = \frac{- a^{2} - x \left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{4} x}+C$$

$$$\int \frac{1}{x^{2} \left(a^{2} + x^{2}\right)}\, dx = \frac{- a^{2} - x \left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{4} x} + C$$$A