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Explanation

`I=int_(-pi)^(pi)(2x(1+sinx))/(1+cos^(2)x) dx`
`=int_(-pi)^(pi)(2x)/(1+cos^(2)x)dx+2int_(-pi)^(pi)(x sin x)/(1+cos^(2)x)`……………1
`=0+4int_(0)^(4)(x sinx)/(1+cos^(2)x)dx`
`=4int_(0)^(pi)((pi-x)sin(pi-x))/(1+cos^(2)(pi-x))dx`
`=4int_(0)(pi)((pi-x)sinx)/(1+cos^(2)x)dx`
`=4pii int_(0)^(pi)(sinx)/(1+cos^(2)x)dx-4 int_(0)^(pi)(x sinx)/(1+cos^(2)x)dx`
or `2I=4pi int_(0)^(pi)(sinx)/(1+cos^(2)x) dx`
or `I=2i int_(0)^(pi)(sinx)/(1+cos^(2)x) dx`
Put `cosx=t` so that `-sinx dx=dt`
when `x=0, t=1`, when `x=pi,t=-1`,
`:.I=2pi int_(1)^(-1)(-dt)/(1+t^(2))=4pi int_(0)^(1)(dt)/(1+t^(2))`
`=4pi [tan^(-1)t]_(0)^(1)`
`=4pi(pi)/4=pi^(2)`