Question

The value of the definite integral ${\int }_1^{\infty }{(e^{x+1}+e^{3-x})}^{-1}dx$ is

• A. $\frac{\pi }{e^2}$
• B. $-\frac{\pi }{e^2}$
• C. $\frac{\pi }{2e^2}$
• D. $-\frac{\pi }{2e^2}$

Answer 1

The correct option is C $\frac{\pi }{2e^2}$

${\int }_1^{\infty }\frac{dx}{e^{x+1}+e^{3-x}}$

taking $e^x=t$

$e^{x}dx=dt$

$dx=\frac{dt}{t}$

$\therefore {\int }_1^{\infty }\frac{dx}{e^{x+1}+e^{3-x}}={\int }_1^{\infty }\frac{dt}{te+e^3}=\frac{1}{e}{\int }_0^{\infty }\frac{dt}{t^2+e^2}$

$=\frac{1}{e}{\left|\frac{1}{e}ta{n}^{-1}t\right|}_0^{\infty }$

$=\frac{1}{e}\times \frac{1}{e}\times \frac{\pi }{2}=\frac{\pi }{2e^2}$

$\therefore$ option $C$ is correct