∫∞0log(x+1x)dx1+x2
Put
x=tanθ∴dx=sec2θdθ
When x=0⇒θ=0
When x=∞⇒θ=π2
∴I=∫π/20log(tanθ+cotθ)dθ
=∫π/20logsin2θ+cos2θsinθcosθdθ
=−∫π/20logsinθcosθdθ
We know that
∫π/20logsinθ=−π2log2
∫π/20logcosθ=−π2log2
=−[∫π/20logsinθdθ+∫π/20logcosθdθ]
=−[−π2log2−π2log2]
I=πlog2.