Question
Evaluate the definite integral $\int\limits_2^3 {\frac{{dx}}{{{x^2} - 1}}} $

Answer

$\int\limits_2^3 {\frac{{dx}}{{{x^2} - 1}}} = \int\limits_2^3 {\frac{1}{{{x^2} - {1^2}}}dx} $

$= \left( {\frac{1}{{2\left( 1 \right)}}\log \left| {\frac{{x - 1}}{{x + 1}}} \right|} \right)_2^3$

$= \frac{1}{2}\log \left| {\frac{{3 - 1}}{{3 + 1}}} \right| - \frac{1}{2}\log \left| {\frac{{2 - 1}}{{2 + 1}}} \right|$

$= \frac{1}{2}\log \left| {\frac{1}{2}} \right| - \frac{1}{2}\log \left| {\frac{1}{3}} \right|$

$= \frac{1}{2}\left( {\log \frac{1}{2} - \log \frac{1}{3}} \right)$

$= \frac{1}{2}\log \frac{{\frac{1}{2}}}{{\frac{1}{3}}}$

$ = \frac{1}{2}\log \frac{3}{2}$

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