分部积分法题目
∫x㏑(x-1)dx
∫x㏑(x-1)dx =∫(1/2)ln(x-1)d(x^2) =(1/2)[x^2*ln(x-1)-∫x^2/(x-1)dx] =(1/2)[x^2*ln(x-1)-∫(x^2-1+1)/(x-1)dx] =(1/2)[x^2*ln(x-1)-∫(x+1)dx-∫1/(x-1)dx] =(1/2)[x^2*ln(x-1)-(1/2)x^2-x-ln(x-1)]+C
答:∫(lnx)^2dx=x(lnx)^2-∫xd(lnx)^2=x(lnx)^2-∫x*2lnx*1/x dx =x(lnx)^2-2∫lnxdx=x(lnx)^...详情>>
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