麻了无从下手

∫arctanx/(1+x) dx = ln(1+x) arctanx - ∫ln(1+x)/(1+x^2) dx = π/4 ln2 - I
其中，令∫ln(1+x)/(1+x^2) dx = I
ln(1+x) = ∫x/(1+xy) dy y[0,1]
∫ln(1+x)/(1+x^2) dx = ∫∫x/(1+xy)(1+x^2) dxdy
令x/(1+xy)(1+x^2) = (Ax+B)/(1+x^2) + C/(1+xy)
(Ax+B)(1+xy) + C(1+x^2) = x
Ax+B + Ay x^2 + Bxy + C+Cx^2-x = 0
(Ay+C)x^2 + (A+ By-1)x + B+ C = 0
B+ C = 0 B=-C
Ay+C = 0 B = Ay
A+ By-1 =0 A+ Ayy=1
A = 1/(1+y^2)
B = y/(1+y^2) C= -y/(1+y^2)
∫∫(Ax+B)/(1+x^2) + C/(1+xy) dxdy
=∫∫(x/(1+y^2)+y/(1+y^2))/(1+x^2) -y/(1+y^2)(1+xy) dxdy
=∫∫x/(1+y^2)(1+x^2) +y/(1+y^2)(1+x^2)-y/(1+y^2)(1+xy) dxdy
由于区间对称性
∫∫y/(1+y^2)(1+xy) dxdy = ∫∫x/(1+y^2)(1+xy) dxdy =I
∫∫x/(1+y^2)(1+x^2)dxdy = ∫∫y/(1+y^2)(1+x^2)dxdy
I=2∫∫x/(1+y^2)(1+x^2)dxdy - I
I = ∫∫x/(1+y^2)(1+x^2)dxd
=1/2 arctany*ln(1+x^2)
=1/2 π/4 ln2
原式 = π/4*ln2 - I = π/4*ln2 - 1/2(π/4*ln2)
= 1/2(π/4*ln2)
=(π*ln2)/8
重积分操作终于完成了。之前直接用arctanx来做成重积分失败，但是分部再重积分终于成功了。