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

求$$$\int \frac{1}{a - x^{2}}\, dx$$$。

令 $$$u=x \sqrt{- \frac{1}{a}}$$$。

則 $$$du=\left(x \sqrt{- \frac{1}{a}}\right)^{\prime }dx = \sqrt{- \frac{1}{a}} dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{\sqrt{- \frac{1}{a}}}$$$。

該積分變為

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

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

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

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

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

回顧一下 $$$u=x \sqrt{- \frac{1}{a}}$$$：

$$\frac{\sqrt{- a} \operatorname{atan}{\left({\color{red}{u}} \right)}}{a} = \frac{\sqrt{- a} \operatorname{atan}{\left({\color{red}{x \sqrt{- \frac{1}{a}}}} \right)}}{a}$$

因此，

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

化簡：

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

加上積分常數：

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

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