First slide

Evaluation of definite integrals

Question

$\int_{- \sqrt{3}}^{\sqrt{3}} \frac{dx}{(1 + e^{x}) (1 + x^{2})}$ is equal to

Moderate

A

$\frac{π}{3}$
B

$\frac{π}{6}$
C

$\frac{π}{4}$
D

$π$

Solution

$\begin{array}{l} I = \int_{- \sqrt{3}}^{\sqrt{3}} \frac{dx}{(1 + e^{x}) (1 + x^{2})} \\ Here, f (x) = \frac{1}{(1 + e^{x}) (1 + x^{2})} \\ \Rightarrow f (- x) = \frac{1}{(1 + e^{- x}) \{1 + (- x)^{2}\}} = \frac{e^{x}}{(1 + e^{x}) (1 + x^{2})} \\ [using property \int_{- a}^{a} f (x) dx = \int_{0}^{a} {f (x) + f (- x)} dx] \end{array}$
$\int_{- \sqrt{3}}^{\sqrt{3}} \frac{dx}{(1 + e^{x}) (1 + x^{2})} = \int_{0}^{\sqrt{3}} \frac{1}{(1 + e^{x}) (1 + x^{2})} + \frac{e^{x}}{(1 + e^{x}) (1 + x^{2})} dx$
so, $I = \int_{0}^{\sqrt{3}} \frac{dx}{1 + x^{2}} = {\tan^{- 1} x|}_{0}^{\sqrt{3}} = \frac{π}{3}$

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