Question

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

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{8}$
• D. $1$

Answer 1

The correct option is B

$\frac{\pi }{4}$

$I={\int }_{-1}^1\frac{dx}{(1+e^x)(1+x^2)}$ ....(i)

Put $x=-t\Rightarrow dx=-dt$

$\therefore I={\int }_1^{-1}\frac{dt}{(1+e^{-t})(1+t^2)}$

$\Rightarrow I={\int }_{-1}^1\frac{\left(e^t\right)dt}{(1+e^t)(1+t^2)}$

$\Rightarrow I={\int }_{-1}^1\frac{\left(e^x\right)dx}{(1+e^x)(1+x^2)}$ ...(ii)

Adding (i) and (ii)

$2I={\int }_{-1}^1\frac{dx}{1+x^2}=2{\int }_0^1\frac{dx}{1+x^2}$

$\Rightarrow I=[{\tan }^{-1}x{]}_0^1=\frac{\pi }{4}$