谁会解这个积分啊 ∫(x²/(1+x²)²) dx

令x=tant,则sint=x/√(1+x²),cost=1/√(1+x²),dx=sec²tdt
于是,原式=∫(tan²t/(sec²t)²)sec²tdt
=∫sin²tdt
=(1/2)∫(1-cos(2t))dt
=(1/2)(t-sin(2t)/2)+C (C是任意常数)
=(t-sint*cost)/2+C
=(arctanx-x/(1+x²))/2+C.