Question

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

Answer 1

$\int \frac{e^{m{\tan }^{-1}x}}{1+x^2}dx$
$\begin{array}{c}\text{Let}m{\tan }^{-1}x=t\\ \Rightarrow {\tan }^{-1}x=\frac{t}{m}\\ \Rightarrow =\frac{1}{1+x^2}dx=\frac{dt}{m}\\ \int \frac{e^{m{\tan }^{-1}x}}{1+x^2}dx\\ =\int e^t\frac{dt}{m}\\ =\frac{1}{m}{e}^t+C\\ \Rightarrow =\frac{1}{m}{e}^{m{\tan }^{-1}x}+C\end{array}$