Integrate (1/((e^(x+1))+(e^(3-x))) dx from x=1 to infinity

Integrate (1/((e^(x+1))+(e^(3-x))) dx from x=1 to infinity?

the answer is pi/(4e^2). Id like to know how to solve. This is a tough one. It involves atan i guess

∫(x = 1 to ∞) dx / [e^(x+1) + e^(3-x)]
= ∫(x = 1 to ∞) dx / {e^2 [e^(x-1) + e^(1-x)]}
= e^(-2) * ∫(x = 1 to ∞) dx / [e^(x-1) + e^(1-x)]

= e^(-2) * ∫(x = 1 to ∞) dx / {e^(-(x-1)) * [e^(2(x-1)) + 1]}
= e^(-2) * ∫(x = 1 to ∞) e^(x-1) dx / [e^(2(x-1)) + 1].

Let z = e^(x-1), dz = e^(x-1) dx. So, we obtain
e^(-2) * ∫(z = 1 to ∞) dz / (z^2 + 1)
= e^(-2) arctan z {for z = 1 to ∞}
= e^(-2) (π/2 - π/4)
= π/(4e^2).

I hope this helps!