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