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Q.
The value of definite integral $\int\limits_1^{\infty}\left(e^{x+1}+e^{3-x}\right)^{-1} d x$ is
Integrals
Solution:
$I=\int\limits_1^{\infty} \frac{1}{\left(e^{x+1}+e^{3-x}\right)} d x=\int\limits_1^{\infty} \frac{1}{e\left(e^x+e^2 \cdot e^{-x}\right)} d x=\int\limits_1^{\infty} \frac{e^x}{e\left(e^{2 x}+e^2\right)} d x$
Let $ e ^{ x }= t$
$\therefore I =\frac{1}{ e } \int\limits_{ e }^{\infty} \frac{1}{ t ^2+ e ^2} dt =\frac{1}{ e ^2}\left(\tan ^{-1} \frac{ t }{ e }\right)_{ e }^{\infty}=\frac{1}{ e ^2}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\frac{\pi}{4 e ^2}$