用分部积分法：
∫x*ln(x-1)dx
=1/2∫xln(x-1)dx^2
=1/2x^2ln(x-1)-1/2∫x^2dln(x-1)
=1/2x^2ln(x-1)-1/2∫x^2/(x-1)dx
=1/2x^2ln(x-1)-1/2∫(x^2-1+1)/(x-1)dx
=1/2x^2ln(x-1)-1/2∫[x+1+1/(x-1)]dx
=1/2x^2ln(x-1)-1/4x^2-x/2-1/2ln(x-1)+C
再问： =1/2x^2ln(x-1)-1/2∫[x+1+1/(x-1)]dx 这一步怎么得来的
再答： 用分部积分法： ∫udv=uv-∫vdu => ∫x*ln(x-1)dx =1/2∫ln(x-1)dx^2 =1/2x^2ln(x-1)-1/2∫x^2dln(x-1) =1/2x^2ln(x-1)-1/2∫x^2/(x-1)dx =1/2x^2ln(x-1)-1/2∫(x^2-1+1)/(x-1)dx =1/2x^2ln(x-1)-1/2∫[x+1+1/(x-1)]dx =1/2x^2ln(x-1)-1/4x^2-x/2-1/2ln(x-1)+C
