Evaluate the following integrals


Let

Let


Let x=tanθ


θ=tan-1x




=sin-1 (2sinθcosθ)


=sin-1 (sin2θ)


Hence f(x)=2θ


=2tan-1x


Hence


Using integration by parts, we get



-(1)


Let


Let 1+x2=t


2x dx=dt.


Also, when x=0, t=1


and when x=1, t=2


Hence,




–(2)


Substituting value of (2) in (1), we get



1