Evaluate e^ ^ -1x 1+x^2 dx - Vidyadip

Question

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

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Answer

Let $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$
Let $\text{I}=\int\frac{\text{e}^{\tan^{-1\text{x}}}}{1+\text{x}^2}\text{ dx}$
$=\int\text{e}^{\text{t}}\text{dt}$
$=\text{e}^{\text{t}}+\text{C}$
$=\text{e}^{\tan^{-1}}\text{x}+\text{C}$

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