Worth mention is the use of partial fractions to derive differential operator representations, viz. $\rm\displaystyle \frac{5-x^2}{(1-x)^3}\ =\ \frac{4}{(1-x)^3}- \frac{2}{(1-x)^2} - \frac{1}{1-x}\ =\ (2\ D^2 + 2\ D - 1)\ \bigg(\frac{1}{1-x}\bigg),\quad D\ =\ \frac{d}{dx} $
Consider a general $2\:$nd-order constant coefficient differential operator applied to $\rm\:1/(1-x)\:.$
$\rm\qquad\displaystyle \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}\ =\ (a\ D^2 + b\ D + c)\ \bigg(\!\!\frac{1}{1-x}\bigg)$ $\rm\displaystyle\phantom{\qquad \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}}\ =\ (a\ D^2 + b\ D + c)\ \sum_{k\:=\:0}^{\infty}\ x^k $ $\rm\displaystyle\phantom{\qquad \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}}\ =\ \sum_{k\:=\:0}^{\infty}\ ((k+2)\:(k+1)\:a+(k+1)\:b+c)\ x^k$
Thus $\rm\ a = 2 = b,\ c = -1\ $ yields
$\rm \frac{5-x^2}{(1-x)^3}\: =\ \sum_{k\:=\:0}^\infty\ (2\:(k+3)\:(k+1)-1)\ x^k $
Hence we find that the coefficient of $\rm\:x^{12}\:$ equals $\ 2\cdot 15\cdot 13 -1\ =\ 389\:.$