We had our final exam yesterday and one of the questions was to find out the value of:

∫∞0lnx1+x2dx

Interestingly enough, using the substitution x=1t we get – −∫10lnx1+x2dx=∫∞1lnx1+x2dxand therefore ∫∞0lnx1+x2dx=0I was curious to know about the theory behind this interesting (surprising even!) example.

Thank you.

**Answer**

When I see an 1+x2 in the denominator it’s tempting to let θ=arctan(x) and dθ=11+x2dx. When you do that here the integral becomes

∫π20ln(tan(θ))dθ

=∫π20ln(sin(θ))dθ−∫π20ln(cos(θ))dθ

The two terms cancel because cos(θ)=sin(π2−θ).

Also, if you do enough of these, you learn that doing the change of variables from x to 1x converts a dx1+x2 into −dx1+x2, so it becomes one of the “tricks of the trade” for integrals with 1+x2 in the denominator. An example: show this trick can be used to show that the following integral is independent of r:

∫∞0dx(1+x2)(1+xr)

**Attribution***Source : Link , Question Author : Amihai Zivan , Answer Author : Zarrax*