x2/(1+x2)2dx的积分为多少
1个回答

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.