Question:
$\frac{e^{\tan ^{-1} x}}{1+x^{2}}$
Solution:
Let $\tan ^{-1} x=t$
$\therefore \frac{1}{1+x^{2}} d x=d t$
$\Rightarrow \int \frac{e^{\tan ^{-1} x}}{1+x^{2}} d x=\int e^{t} d t$
$=e^{\prime}+\mathrm{C}$
$=e^{\tan ^{-1} x}+\mathrm{C}$
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