Question

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

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

Answer 1

Answer: (A)

$\frac{\pi }{4e^2}$

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

$I={\int }_1^{\infty }\frac{e^x}{e^{2x+1}+e^3}dx=\frac{1}{e}{\int }_1^{\infty }\frac{e^x}{e^{2x}+e^2}dx$

Substitute $e^x=t\Rightarrow e^{x}dx=dt$ and limits will be

Lower limit $x=1\Rightarrow t=e$

Upper limit $x=\infty \Rightarrow t=\infty$

$I=\frac{1}{e}{\int }_e^{\infty }\frac{1}{t^2+e^2}dt$

$=\frac{1}{e}{\left|\frac{1}{e}{\tan }^{-1}\left|\frac{t}{e}\right|\right|}_e^{\infty }=\frac{1}{e^2}|\frac{\pi }{2}-\frac{\pi }{4}|=\frac{\pi }{4e^2}$

Hence, $\left(A\right)$