Evaluate ∫0∞ tan−1⁡xx1+x2dx

I=∫0∞ tan−1⁡xx1+x2dx Put x=tan⁡t so that dx=sec2⁡tdt. Then, I=∫0π/2 ttan⁡t⋅sec2⁡t⋅sec2⁡tdt=∫0π/2 tcot⁡tdt=[tlog⁡sin⁡t]0π/2−∫0π/2 1⋅log⁡sin⁡tdt=−∫0π/2 log⁡sin⁡tdt=−−π2log⁡2=π2log⁡2