Question
Integrate the function: $\frac{e^{\tan ^{-1} x}}{1+x^{2}}$

Answer

let $\tan^{-1} x = t$
$\Rightarrow \frac{1}{1+x^{2}} d x=d t$
$\Rightarrow \int \frac{e^{t an^{-1} x}}{1+x^{2}} d x=\int e^{t} d t$
$= e^t + c$
$=e^{\tan ^{-1} x}+C$

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