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Mathematics 19 Online
OpenStudy (anonymous):

\[\int\limits_{1}^{2}1/(x(x+1)) dx\]

OpenStudy (nowhereman):

You can transform the integrand: \[\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}\]

OpenStudy (anonymous):

\[\int\limits_{1}^{2}dx/x-\int\limits_{1}^{2}dx/(x+1)=\ln(x/(x+1))|_{1}^{2}=\ln(1/2)-\ln(2/3)\]

OpenStudy (anonymous):

ln(1/2)-ln(2/3)=ln(3/4)

OpenStudy (nowhereman):

Shouldn't it be the other way around? \[\ln(2/3) - \ln(1/2) = \ln(4/3)\]

OpenStudy (anonymous):

Yh I got ln(4/3) too

OpenStudy (anonymous):

yes i did a mistake with a minus

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