Question

Evaluate: $\int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx \\ \mathrm{Let}{\tan }^{-1}x=t \\ \Rightarrow \frac{dx}{1+x^2}=dt \\ \therefore I=\int e^{t}dt \\ =e^t+C \\ =e^{{\tan }^{-1}x}+C \end{gathered}$$