方法一:
∫[x^2/(1+x^2)^2]dx
=(1/2)∫[(1+x^2)/(1+x^2)^2]dx-(1/2)∫[(1-x^2)/(1+x^2)^2]dx
=(1/2)∫[1/(1+x^2)]dx-(1/2)∫[(1+x^2-2x^2)/(1+x^2)^2]dx
=(1/2)arctanx-(1/2)∫{[x′(1+x^2)-x(1+x^2)′]/(1+x^2)^2}dx
=(1/2)arctanx-(1/2)∫d[x/(1+x^2)]
=(1/2)arctanx-(1/2)[x/(1+x^2)]+C
=(1/2)arctanx-x/(2+2x^2)+C.
方法二:
令x=tant,则:t=arctanx,dx=(sect)^2dt,于是:
∫[x^2/(1+x^2)^2]dx
=∫{(tant)^2/[1+(tant)^2]^2}(sect)^2dt
=∫{(tant)^2/[1/(sect)^2]}dt
=∫(sint)^2dt
=(1/2)∫(1-cos2t)dt
=(1/2)∫dt-(1/4)∫cos2td(2t)
=(1/2)t-(1/4)sin2t+C
=(1/2)arctanx-(1/2)sintcost+C
=(1/2)arctanx-(1/2)sintcost/[(cost)^2+(sint)^2]+C
=(1/2)arctanx-(1/2)tanx/[1+(tanx)^2]+C
=(1/2)arctanx-(1/2)x/(1+x^2)+C
=(1/2)arctanx-x/(2+2x^2)+C.