Evaluate the following integrals:
Question:

Evaluate: $\int \frac{\mathrm{e}^{\tan ^{-1}}}{1+\mathrm{x}^{2}} \mathrm{dx}$

Solution:

Given, $\int \frac{e^{\tan ^{-1}}}{1+x^{2}} d x$

Let $\tan ^{-1} x=t$

$\partial \frac{d y}{d x}\left(\operatorname{Tan}^{-1} x\right)=\mathrm{dt}$

$\partial \frac{1}{1+x^{2}} d x=d t$

Now, $\int \frac{e^{\tan ^{-1}}}{1+x^{2}} d x$

$=\int e^{t} d t$

$=e^{t}+c$

$=e^{\tan ^{-1} x+c}$

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