Question

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

• A. $\pi /2$
• B. $\pi /4$
• C. $\pi /8$
• D. $\pi /16$

Answer 1

Answer: (B)

$I=\underset{-1}{\overset{1}{\int }}\frac{dx}{\left(1+e^x\right)\left(1+x^2\right)}$ ....(1)
$=\underset{-1}{\overset{1}{\int }}\frac{dx}{1+e^{-x}}\cdot \frac{1}{1+x^2}$(using King)
$I=\underset{-1}{\overset{1}{\int }}\frac{e^{x}dx}{\left(1+e^x\right)\left(1+x^2\right)}$....(2)
adding (1) and (2)
$2I=\underset{-1}{\overset{1}{\int }}\frac{\left(1+e^x\right)dx}{\left(1+e^x\right)\left(1+x^2\right)}=\underset{-1}{\overset{1}{\int }}\frac{dx}{\left(1+x^2\right)}=2\underset{0}{\overset{1}{\int }}\frac{dx}{\left(1+x^2\right)}$
$I=\underset{0}{\overset{1}{\int }}\frac{dx}{\left(1+x^2\right)}={\tan }^{-1}(1)=\pi /4$
[convert it into value of definite integral ' $T$ ' is same as ]