You might want to aim at partial fraction decomposition
$\frac{1}{{\left( {1 - 2x} \right)\left( {1 + 3{x^2}} \right)}} = \frac{A}{{1 - 2x}} + \frac{{Bx + C}}{{1 + 3{x^2}}}$
and then find the coefficients separately.
After equating coefficients, you should get the system $\eqalign{ & A + C = 1 \cr & B - 2C = 0 \cr & 3A - 2B = 0 \cr} $ from where $\eqalign{ & A = 4/7 \cr & B = 6/7 \cr & C = 3/7 \cr} $
$\frac{1}{{\left( {1 - 2x} \right)\left( {1 + 3{x^2}} \right)}} = \frac{1}{7}\left( {\frac{4}{{1 - 2x}} + \frac{{6x + 3}}{{1 + 3{x^2}}}} \right)$
now use the usual expansions
$\eqalign{ & \frac{4}{{1 - 2x}} = 4\sum\limits_{n = 0}^\infty {{2^n}{x^n}} \cr & \frac{1}{{1 + 3{x^2}}} = \sum\limits_{n = 0}^\infty (-1)^n {{{ { 3} }^{n}}{x^{2n}}} \cr} $
You might want to consider odd and even terms separately.