Evaluate the following integrals: e^ m ^ -1 x 1+x^2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$

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Answer

$\int\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$

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

$\Rightarrow\Big(\frac{1}{1+\text{x}^2}\Big)\text{dx}=\text{dt}$

Now, $\int\Big(\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{1+\text{x}^2}\Big)\text{ dx}$

$=\int\text{e}^{\text{mt}}\text{dt}$

$=\frac{\text{e}^{\text{mt}}}{\text{m}}+\text{C}$

$=\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{\text{m}}+\text{C}$

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