can anyone tell me what the the value of this is - i think it is ln(1/3) but my book says ln(3)

$\displaystyle

\int\limits_{ - \infty }^2 {[\frac{1}{{x - 1}}} - \frac{1}{{x + 1}}]dx

$

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- Feb 6th 2009, 05:05 PMbillaEasy improper integral
can anyone tell me what the the value of this is - i think it is ln(1/3) but my book says ln(3)

$\displaystyle

\int\limits_{ - \infty }^2 {[\frac{1}{{x - 1}}} - \frac{1}{{x + 1}}]dx

$ - Feb 6th 2009, 05:35 PMKrizalid
$\displaystyle \left. \ln \left| \frac{x-1}{x+1} \right| \right|_{-\infty }^{2}=\ln \left( \frac{1}{3} \right)=-\ln 3.$

- Feb 6th 2009, 05:36 PMbilla
woot book is wrong i am right

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