Question

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

Answer 1

`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`