>>566
次にＸで積分　→　∫[0,1] x(x*arctanx -(1/2)log(1+x^2) )dx

第一項を部分積分　→((1/3)x^3*arctanx)_[0,1] -∫[0,1](1/3)x^3/(1+x^2) dx
=(1/3)arctan(1) -(1/3)∫[0,1] (x-x/(1+x^2) )dx
=π/12 - (1/3)(1/2-log2) =π/12 -1/6 +(1/6)log2　←∫ x/(1+x^2) dx =(1/2)log(1+x^2)を使う

第二項を積分　→(-1/2)∫xlog(1+x^2) =(-1/2) ((1/2)x^2*log(1+x^2)_[0,1] -(-1/2)∫(1/2)x^2*(2x)/(1+x^2) dx
=(-1/4)log2 -(-1/2)2∫[0,1] x^3/(1+x^2) dx =(-1/4)log2 -(-1/2)(1/2-(1/2)log2)
=(-1/4)log2 +1/4 -(1/4)log2 =(-1/2)log2+1/4

よって π/12-1/6 +(1/6)log2 -(1/2)log2 +1/4 = π/12 -(1/3)log2 +1/12