Question

Integrate the following functions.
$\int \frac{e^{ta{n}^{-1}x}}{1+x^2}dx.$

Answer 1

$\int \frac{e^{ta{n}^{-1}x}}{1+x^2}dx$
Let $ta{n}^{-1}x=t\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\Rightarrow dx=(1+x^2)dt$
$\therefore \int \frac{e^{ta{n}^{-1}x}}{1+x^2}dx=\int \frac{e^t}{1+x^2}(1+x^2)dt=\int e^{t}dt=e^t+C=e^{ta{n}^{-1}x}+C$