∫ [∞,-∞] e^x/1+e^2x dx

Answer 1

let u = e^x, du = e^x dx
and you have
∫ [∞,-∞] 1/(1+u^2) du = arctan(u) [∞,-∞] = π

Answer 2

I thought arctan(∞)=π/2
Who did you get π

Answer 3

you are correct.
and what is arctan(-∞) ?

Answer 4

Oh π\2-(-π/2)=π
Got it thanks