Answer
$\dfrac{1}{9}-\dfrac{1}{x}+\dfrac{25}{18(3x+2)}+\dfrac{29}{18(3x-2)} $
Work Step by Step
We will write the partial decomposition as:
$\dfrac{1}{9}+\dfrac{\dfrac{4}{9}x+4}{x(3x+2) (3x-2)}=\dfrac{1}{9}+\dfrac{A}{x}+\dfrac{B}{3x+2}+\dfrac{C}{3x-2} ~~~~(a)$
This implies that $\dfrac{4}{9}x+4=A(3x+2)(3x-2)+Bx(3x-2)+Cx(3x-2)$
Plug $x=0, -1$ to obtain: $4=-4A \implies A=-1$
After solving these equations, we get: $B=\dfrac{25}{18} $ and $C=\dfrac{29}{18}$
So, the equation (a) becomes:
$\dfrac{1}{9}+ \dfrac{\dfrac{4}{9}x+4}{x(3x+2) (3x-2)}=\dfrac{1}{9}-\dfrac{1}{x}+\dfrac{25}{18(3x+2)}+\dfrac{29}{18(3x-2)} $
