1. ## partial fractions

Could someone please explain how to solve this?

2. Originally Posted by Haris

Could someone please explain how to solve this?
Take out a factor of a quarter and you get:

$\displaystyle \frac{1}{4}.\frac{1}{x^2-\frac{9}{4}}$

This is a difference of two squares:

$\displaystyle \frac{1}{4}.\frac{1}{x^2 -(\frac{3}{2})^2} = \frac{1} {4}.\frac{1}{(x-\frac{3}{2})(x+\frac{3}{2})}$

Is that better?

3. Originally Posted by Haris

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 $\displaystyle A$.

4. $\displaystyle (2x+3)-(2x-3)=6.$

5. Originally Posted by Krizalid
$\displaystyle (2x+3)-(2x-3)=6.$
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?

6. Yeah.