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Q.
The value of the 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{d x}{\left(e \cdot e^x+e^3 \cdot e^{-x}\right)}=\int\limits_1^{\infty} \frac{e^x d x}{e\left(e^{2 x}+e^2\right)} $ (multiply $N^x$ and $D^x$ by $\left.e^x\right)$
put $e ^{ x }= t \Rightarrow e ^{ x } dx = dt$
$I =\frac{1}{ e } \int\limits_{ e }^{\infty} \frac{ dt }{ t ^2+ e ^2}=\left.\frac{1}{ e ^2} \tan ^{-1} \frac{ t }{ e }\right|_e ^{\infty}=\frac{1}{ e ^2}\left[\frac{\pi}{2}-\frac{\pi}{4}\right]=\frac{\pi}{4 e ^2}$