$\cfrac{1}{{{x^2} - 9}}$

Let $I = \int {\cfrac{{dx}}{{{x^2} - 9}}}$

Let $\cfrac{1}{{{x^2} - 9}} = \cfrac{1}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \cfrac{A}{{x - 3}} + \cfrac{B}{{x + 3}}$

$\Rightarrow$ $1 = A\left( {x + 3} \right) + \left( {x - 3} \right)$

….(i)
Putting $x = 3$ in (i),

we get $1 = A\left( {3 + 3} \right)$ $\Rightarrow$ $A = \cfrac{1}{6}$

Putting $x = - 3$ in (i),

we get $1 = B\left( { - 3 - 3} \right)$ $\Rightarrow$ $B = - \cfrac{1}{6}$

$\therefore$ $I = \int {\cfrac{{dx}}{{{x^2} - 9}}} = \cfrac{1}{6}\int {\left[ {\cfrac{1}{{x - 3}} - \cfrac{1}{{x + 3}}} \right]} dx$

$= \cfrac{1}{6}\left[ {\log \left| {x - 3} \right| - \log \left| {x + 3} \right|} \right] + C = \cfrac{1}{6}\log \left| {\cfrac{{x - 3}}{{x + 3}}} \right| + C$