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

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

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

则$$$du=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x^{2} + 1}$$$ (步骤见»)，并有$$$\frac{dx}{x^{2} + 1} = du$$$。

该积分可以改写为

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：

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

回忆一下 $$$u=\operatorname{atan}{\left(x \right)}$$$:

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

因此，

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

加上积分常数：

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

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