求不定积分∫[x·arctanx／﹙1＋x^2)]dx,过程，谢谢

用三角函数法解：
令y=arctanx，dy=1/(1+x²) dx
∫ xarctanx/(1+x²)² dx
= ∫ xarctanx/(1+x²) * 1/(1+x²) dx
= ∫ ytany/(1+tan²y) dy
= ∫ ytanycos²y dy
= ∫ y*siny/cosy*cos²y dy
= ∫ ysinycosy dy
= (1/2)∫ ysin2y dy
= (1/2)(-1/2)∫ yd(cos2y)
= (-1/4)ycos2y + (1/4)∫ cos2y dy
= (-1/4)ycos2y + (1/4)(1/2)sin2y + C
= (1/8)sin2y - (1/4)ycos2y + C
= [x+(x²-1)arctanx]/[4(1+x²)] + C

∫ [ x·arctanx／(1＋x^2)^2] dx 换元 u=arctanx, x=tanu, dx=(secu)^2 du
= ∫ tanu * u / (secu)^2 du = (1/2) ∫ u sin(2u) du
= (-1/4) ∫ u d cos(2u) 分部积分
= (-1/4) [ u cos(2u) - (1/2)sin(2u) ] + C
= (-1/4) arctanx * (1-x²)/(1+x²) + (1/4) x /(1+x²) + C

∫[x·arctanx／﹙1＋x^2)^2]dx
=1/2∫arctanxd1/(1+x^2)
=1/2arctanx/(1+x^2)-1/2∫1/(1+x^2)darctanx
=1/2arctanx/(1+x^2)- 1/4∫1/(1+x^2)^2dx
对于后面的一项不定积分，应用有理式不定积分解决。