Could someone please explain how to solve this?
note that $\displaystyle \frac 1{4x^2 - 9} = \frac 1{(2x + 3)(2x - 3)} = \frac A{2x + 3} + \frac B{2x - 3}$
Now multiply through by $\displaystyle (2x + 3)(2x - 3)$, you get
$\displaystyle 1 = A(2x - 3) + B(2x + 3)$ ................(1)
now plug in $\displaystyle x = \frac 32$ and solve for $\displaystyle B$, then set $\displaystyle x = - \frac 32$ and solve for $\displaystyle A$.
haha, ok, fine Kriz
without partial fractions: $\displaystyle \frac 1{(2x + 3)(2x - 3)} = \frac 16 \cdot \frac 6{(2x + 3)(2x - 3)} = \frac 16 \cdot \frac {2x + 3 - (2x - 3)}{(2x + 3)(2x - 3)} = $ $\displaystyle \frac 16 \left( \frac 1{2x - 3} - \frac 1{2x + 3} \right)$
happy now?