问题：利用分部积分法求∫x ln(x-1)dx的不定积分． 答案： ∫x ln(x-1)dx=x^2/2* ln(x-1)-∫x^2/2ln(x-1)’dx =x^2/2* ln(x-1)-∫x^2/2(x-1)dx =x^2/2* ln(x-1)-∫(x^2-x)/2(x-1)dx-∫x/2(x-1)dx =x^2/2* ln(x-1)-∫x/2dx-∫x/2(x-1)dx =x^2/2* ln(x-1)-x^2/4-∫x/2(x-1)dx =x^2/2* ln(x-1)-x^2/4-∫(x-1)/2(x-1)dx-∫1/2(x-1)dx =x^2/2* ln(x-1)-x^2/4-∫1/2dx-∫1/2(x-1)d(x-1) =x^2/2* ln(x-1)-x^2/4-x/2-∫1/2(x-1)d(x-1) =x^2/2* ln(x-1)-x^2/4-x/2-ln(x-1)/2 参考文献：iask.sina.com.cn/b/8960613.html