# Easy improper integral

• Feb 6th 2009, 05:05 PM
billa
Easy 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)

$
\int\limits_{ - \infty }^2 {[\frac{1}{{x - 1}}} - \frac{1}{{x + 1}}]dx
$
• Feb 6th 2009, 05:35 PM
Krizalid
$\left. \ln \left| \frac{x-1}{x+1} \right| \right|_{-\infty }^{2}=\ln \left( \frac{1}{3} \right)=-\ln 3.$
• Feb 6th 2009, 05:36 PM
billa
woot book is wrong i am right

ty