Question

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

• A. $\frac{\pi }{4c^2}$
• 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)

$I=\underset{1}{\overset{\infty }{\int }}\frac{dx}{\left(e\cdot e^x+e^3\cdot e^{-x}\right)}=\underset{1}{\overset{\infty }{\int }}\frac{e^{x}dx}{e\left(e^{2x}+e^2\right)}$ (multiply $N^x$ and $D^x$ by $e^x)$
put $e^x=t\Rightarrow e^{x}dx=dt$
$I=\frac{1}{e}\underset{e}{\overset{\infty }{\int }}\frac{dt}{t^2+e^2}={\frac{1}{e^2}{\tan }^{-1}\frac{t}{e}|}_e^{\infty }=\frac{1}{e^2}\left[\frac{\pi }{2}-\frac{\pi }{4}\right]=\frac{\pi }{4e^2}$