1/x(x+1)=1/x-1/(x+1)
所以原式=∫[(ln(x+1)-lnx]*[1/x-1/(x+1)]dx
=∫[(ln(x+1)-lnx]d[lnx-(ln(x+1)]
=-∫[lnx-ln(x+1)]d[lnx-(ln(x+1)]
=-(1/2)*[lnx-(ln(x+1)]^2+C
=-(1/2)[lnx/(x+1)]^2+C
或者因为lnx-(ln(x+1)=-[(ln(x+1)-lnx]
所以-(1/2)*[lnx-(ln(x+1)]^2+C
=-(1/2)*[(ln(x+1)-lnx]^2+C
=-(1/2)*[(ln(x+1)/x]^2+C
两者一样