If $\frac{1}{(1-2\mathrm{x})(1+3\mathrm{x})}$ is to be expanded as a power series of $\mathrm{x},$ then

$\begin{array}{rcl}\frac{1}{(1-2\mathrm{x})(1+3\mathrm{x})}& =& \frac{{\displaystyle \frac{2}{5}}}{1-2\mathrm{x}}+\frac{{\displaystyle \frac{3}{5}}}{1+3\mathrm{x}}\\ & =& \frac{2}{5} {\left(1-2\mathrm{x}\right)}^{-1}+\frac{3}{5}{\left(1+3\mathrm{x}\right)}^{-1}\end{array}$

it can be expanded as powers of x if 

if  $\begin{array}{rcl}\left|2\mathrm{x}\right|& <& 1, \left|3\mathrm{x}\right|<1\end{array}$

     $\begin{array}{rcl}\left|\mathrm{x}\right|& <& \frac{1}{2} \left|\mathrm{x}\right|<\frac{1}{3}\\ \frac{-1}{2}& <& \mathrm{x}<\frac{1}{2} \& \frac{-1}{3}<\mathrm{x}<\frac{1}{3}\end{array}$

Common Part $\frac{-1}{3}<\mathrm{x}<\frac{1}{3}$