$$$\frac{e^{\operatorname{atan}{\left(x \right)}}}{x^{2} + 1}$$$ 的積分
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求$$$\int \frac{e^{\operatorname{atan}{\left(x \right)}}}{x^{2} + 1}\, dx$$$。
解答
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 $$$\frac{dx}{x^{2} + 1} = du$$$。
因此，
$${\color{red}{\int{\frac{e^{\operatorname{atan}{\left(x \right)}}}{x^{2} + 1} d x}}} = {\color{red}{\int{e^{u} d u}}}$$
指數函數的積分為 $$$\int{e^{u} d u} = e^{u}$$$：
$${\color{red}{\int{e^{u} d u}}} = {\color{red}{e^{u}}}$$
回顧一下 $$$u=\operatorname{atan}{\left(x \right)}$$$：
$$e^{{\color{red}{u}}} = e^{{\color{red}{\operatorname{atan}{\left(x \right)}}}}$$
因此，
$$\int{\frac{e^{\operatorname{atan}{\left(x \right)}}}{x^{2} + 1} d x} = e^{\operatorname{atan}{\left(x \right)}}$$
加上積分常數：
$$\int{\frac{e^{\operatorname{atan}{\left(x \right)}}}{x^{2} + 1} d x} = e^{\operatorname{atan}{\left(x \right)}}+C$$
答案
$$$\int \frac{e^{\operatorname{atan}{\left(x \right)}}}{x^{2} + 1}\, dx = e^{\operatorname{atan}{\left(x \right)}} + C$$$A