设arctanx=t,x=tant,dx=(sect)^2dt
∫x*arctanx/(1+x^2)^2dx=∫t * tant /(sect)^4 *(sect)^2dt=∫t * tant * (cost)^2 dt=∫t * sint * cost dt=(1/2)∫t * sin2t dt,分部积分后得(-1/4)tcos2t+(1/8)sin2t+C,再用万能公式把cos2t=(1-x^2)/(1+x^2),sin2t=2x/(1+x^2),t=arctanx代入即可.