∫xlnxdx/(1-x^2)^2

∫xlnxdx/(1-x^2)^2
1/(1-x^2)'=2x/(1-x^2)^2
所以:
原式=1/2∫lnxd(1/(1-x^2))
=1/2[(lnx/(1-x^2)-∫1/(1-x^2)d(lnx)]
=1/2[lnx/(1-x^2)-∫1/(x(1-x^2))dx]
1/(x(1-x^2))=1/(x(x-1)(x+1))=1/x(1/(1-x)+1/(1+x))/2
=1/(x(1-x))/2+1/(x(1+x))/2
=1/2[1/x+1/(1-x)+1/x-1/(1+x)]
所以)∫1/(x(1-x^2))dx
=1/2∫[1/x+1/(1-x)+1/x-1/(1+x)]dx
=1/2[lnx-ln(1-x)+lnx-ln(1+x)]
=lnx-1/2ln(1-x^2)
原式=1/2[lnx/(1-x^2)-(lnx-1/2ln(1-x^2))]