What is int_-oo^oo (e^x)/((e^(2x))+3) dx?

1 Answer
Mar 24, 2016

pi/sqrt3

Explanation:

int_-oo^oo e^x/(e^(2x)+3)dx=

Using trigonometric substitution
e^x=sqrt3tany
e^x*dx=sqrt3sec^2y*dy

Since tany=e^x/sqrt3, when
x->-oo => tany -> -oo => y->(-pi/2)^"+"
x->oo => tany -> oo => y->(pi/2)^"-"

So the expression becomes

=int_((-pi/2)^"+")^((pi/2)^"-") (sqrt3sec^2y)/(3tan^2y+3)dy=int_((-pi/2)^"+")^((pi/2)^"-") (sqrt3cancel(sec^2y))/(3cancel(sec^2y))dy
=(y/sqrt3)|_((-pi/2)^"+")^((pi/2)^"-")
=1/sqrt3[pi/2-(-pi/2)]=pi/sqrt3