被积分的函数是xln(x-1)?
=x^2ln(x-1)+∫x(ln(x-1)+x/(x-1))dx=x^2ln(x-1)+∫xln(x-1)dx+∫x^2/(x-1)dx;
所以, 2∫xln(x-1)dx=x^2ln(x-1)-∫(x+1)dx+∫1/(x-1)dx, ∫xln(x-1)dx=x^2ln(x-1)/2-((x+1)^2)/2-ln(x-1)+C
用分步积分
∫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
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