Question

The value of definite integral $\underset{1}{\overset{\infty }{\int }}{\left(e^{x+1}+e^{3-x}\right)}^{-1}dx$ is

• A. $\frac{\pi }{4e^2}$
• B. $\frac{\pi }{4e}$
• C. $\frac{1}{e^2}\left(\frac{\pi }{2}-{\tan }^{-1}\frac{1}{e}\right)$
• D. $\frac{\pi }{2e^2}$

Answer 1

Answer: (A)

$I=\underset{1}{\overset{\infty }{\int }}\frac{1}{\left(e^{x+1}+e^{3-x}\right)}dx=\underset{1}{\overset{\infty }{\int }}\frac{1}{e\left(e^x+e^2\cdot e^{-x}\right)}dx=\underset{1}{\overset{\infty }{\int }}\frac{e^x}{e\left(e^{2x}+e^2\right)}dx$
Let $e^x=t$
$\therefore I=\frac{1}{e}\underset{e}{\overset{\infty }{\int }}\frac{1}{t^2+e^2}dt=\frac{1}{e^2}{\left({\tan }^{-1}\frac{t}{e}\right)}_e^{\infty }=\frac{1}{e^2}\left(\frac{\pi }{2}-\frac{\pi }{4}\right)=\frac{\pi }{4e^2}$