答案肯定不对,ln后面不可能是x,而是x-1
原式=-∫ln(x-1)d(1/x)
=-ln(x-1)/x+∫(1/x)dln(x-1)
=-ln(x-1)/x+∫1/x(x-1)dx
=-ln(x-1)/x+∫[1/x-1/(x-1)]dx
=-ln(x-1)/x+lnx-ln(x-1)+C
答案肯定不对,ln后面不可能是x,而是x-1
原式=-∫ln(x-1)d(1/x)
=-ln(x-1)/x+∫(1/x)dln(x-1)
=-ln(x-1)/x+∫1/x(x-1)dx
=-ln(x-1)/x+∫[1/x-1/(x-1)]dx
=-ln(x-1)/x+lnx-ln(x-1)+C