Evaluate the following integral ∫_0^∞(x tan^(-1)⁡x)/(1+x^2 )^2 dx

∫_0^∞(x tan^(-1)⁡x)/(1+x^2 )^2  dx
Let  x=tan⁡θ→dx=〖sec〗^2⁡θ dθ,x=0→θ=0,x=∞→θ=π/2
∴I=∫_0^(π/2)(θ tan⁡θ)/(1+〖tan〗^2 θ)^2 .〖sec〗^2 θdθ=∫_0^(π/2)〖θ tan⁡θ 〗.〖cos〗^2 θdθ
=∫_0^(π/2)〖θ sin⁡θ 〗.cos⁡θ dθ=1/2 ∫_0^(π/2)〖θ sin⁡2θ 〗 dθ
By parts:
u=θ                      dv=sin⁡2θ dθ
du=dθ                  v=-1/2  cos⁡2θ
∴I=1/2 [├ -1/2 θ cos⁡θ ┤|_0^(π/2)+├ 1/4  sin⁡2θ ┤|_0^(π/2) ]=1/4 [π/2]=π/8