<s>[INST] How to compute $\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}}$? How to compute $\int_{0}^{+\infty} \frac{dx}{e^{x+1}  + e^{3-x}}$?
My partial solution: 
$$
\int_{0}^{+\infty} \frac{dx}{e^{x+1}  + e^{3-x}} = \int_{0}^{+\infty} \frac{dx}{e^{3-x}(1  + e^{2x-2})} \\
= \int_{0}^{+\infty} \frac{e^{x-3}dx}{1  + e^{2x-2}}.
$$
Thank you very much.
 [/INST] It is natural to let $y=e^x$. Then $dy=e^x\,dx$, so $dx=\frac{1}{y}\,dy$. Thus we want
$$\int_1^\infty \frac{1}{y}\cdot \frac{1}{e\cdot y+e^3 \cdot \frac{1}{y}}\,dy,$$
which simplifies to
$$\int_1^\infty \frac{1}{e}\cdot\frac{1}{y^2+e^2}\,dy.$$
Let $y=et$, and we are nearly finished.
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