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

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

Let $$$u=\operatorname{atan}{\left(x \right)}$$$.

Then $$$du=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x^{2} + 1}$$$ (steps can be seen »), and we have that $$$\frac{dx}{x^{2} + 1} = du$$$.

The integral can be rewritten as

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

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

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

Recall that $$$u=\operatorname{atan}{\left(x \right)}$$$:

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

Therefore,

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

Add the constant of integration:

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

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