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

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

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$$$\frac{dx}{x^{2} + 1} = du$$$。

因此，

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

指数函数的积分为 $$$\int{e^{u} d u} = e^{u}$$$：

$${\color{red}{\int{e^{u} d u}}} = {\color{red}{e^{u}}}$$

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

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

因此，

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

加上积分常数：

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

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