求∫[(ln(x+1)-lnx)/(x(x+1)) - 知识问答

求∫[(ln(x+1)-lnx)/(x(x+1))]dx

3个回答

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(...

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