Question

The value of the definite integral $\underset{-1}{\overset{1}{\int }}\frac{dx}{(1+e^x)(1+x^2)}$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{8}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{16}$

Answer 1

The correct option is A $\frac{\pi }{4}$

$I=\underset{-1}{\overset{1}{\int }}\frac{dx}{(1+e^x)(1+x^2)}\cdots \left(i\right)$
Using the property: $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$=\underset{-1}{\overset{1}{\int }}\frac{dx}{1+e^{-x}}.\frac{1}{1+x^2}$
$I=\underset{-1}{\overset{1}{\int }}\frac{e^{x}dx}{(1+e^x)(1+x^2)}\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right),$ we get
$2I=\underset{-1}{\overset{1}{\int }}\frac{(1+e^x)dx}{(1+e^x)(1+x^2)}$
$=\underset{-1}{\overset{1}{\int }}\frac{dx}{(1+x^2)}$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
$=2\underset{0}{\overset{1}{\int }}\frac{dx}{(1+x^2)}$
$I=\underset{0}{\overset{1}{\int }}\frac{dx}{(1+x^2)}={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$