Question:
Evaluate the following integrals:
$\int \frac{e^{m \tan ^{-1} x}}{1+x^{2}} d x$
Solution:
Assume $\tan ^{-1} x=t$
$d\left(\tan ^{-1} x\right)=d t$
$\Rightarrow \frac{1}{x^{2}+1}=d t$
Substituting $\mathrm{t}$ and $\mathrm{dt}$
$\Rightarrow \int e^{m t} d t$
$\Rightarrow \frac{e^{m t}}{m}+c$
But $t=\tan ^{-1}_{x}$
$\Rightarrow \frac{\mathrm{e}^{\mathrm{mtan}^{-1} \mathrm{x}}}{\mathrm{m}}+\mathrm{c}$
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