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

Answer

Let $\text{I}=\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}$
Let $\tan\text{e}^{\text{x}}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\text{dx = dt}$
So, $\text{I}=\int\frac{\text{dt}}{1+\text{t}^{2}}$
$=\tan^{-1}(\text{t})+\text{C}$ $\big[\text{Since,}\int\frac{1}{1+\text{x}^2}\text{dx}=\tan^{-1}\text{x+c}\big]$
$\text{I}=\tan^{-1}(\text{e}^{\text{x}})+\text{C}$

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