Question

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

Answer 1

$$\begin{gathered} \int \left(\frac{e^{m{\tan }^{-1}x}}{1+x^2}\right)dx \\ \mathrm{Let}{\tan }^{-1}x=t \\ \Rightarrow \left(\frac{1}{1+x^2}\right)dx=dt \\ \mathrm{Now},\int \left(\frac{e^{m{\tan }^{-1}x}}{1+x^2}\right)dx \\ =\int e^{mt}dt \\ =\frac{e^{mt}}{m}+\mathrm{C} \\ =\frac{e^{m{\tan }^{-1}x}}{m}+\mathrm{C} \end{gathered}$$