$\int_{- 1}^{1} \frac{1}{(1 + e^{x})} \times \frac{1}{(1 + x^{2})} d x$

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Solution

We have,
$\begin{matrix} \int_{- 1}^{1} \frac{1}{(1 + e^{x})} \times \frac{1}{(1 + x^{2})} d x I = \int_{- 1}^{1} \frac{e^{x}}{1 + e^{x}} \times \frac{1}{1 + x^{2}} d x 2 I = \int_{- 1}^{1} \frac{d x}{1 + x^{2}} 2 I = 2 \int_{0}^{1} \frac{d x}{1 + x^{2}} I = {[{tan}^{- 1} x]}_{0}^{1} = \frac{π}{4} \end{matrix}$
Hence,
we get the values is $\frac{π}{4}$

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