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\[\int\limits_{1}^{2}1/(x(x+1)) dx\]
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You can transform the integrand: \[\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}\]
\[\int\limits_{1}^{2}dx/x-\int\limits_{1}^{2}dx/(x+1)=\ln(x/(x+1))|_{1}^{2}=\ln(1/2)-\ln(2/3)\]
ln(1/2)-ln(2/3)=ln(3/4)
Shouldn't it be the other way around? \[\ln(2/3) - \ln(1/2) = \ln(4/3)\]
Yh I got ln(4/3) too
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yes i did a mistake with a minus
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