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

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

令 $$$u=\operatorname{atan}{\left(x \right)}$$$。

則 $$$du=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x^{2} + 1}$$$ (步驟見»)，並可得 $$$\frac{dx}{x^{2} + 1} = du$$$。

該積分變為

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

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

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

回顧一下 $$$u=\operatorname{atan}{\left(x \right)}$$$：

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

因此，

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

加上積分常數：

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

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