$\begin{align} & \text{Express }\frac{1}{(x-1)(x+2)}\text{ as sum of two partial fractions}\text{. Hence, write down the } \\ & \text{expansion of fraction as a series when }x\text{ is small, up to the term in }{{x}^{2}}.\text{ } \\ & \text{For values of }x\text{ is the expansion valid?} \\ \end{align}$
$\begin{align} & \frac{1}{(x-1)(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x+2)} \\ & \therefore A(x+2)+B(x-1)=1 \\ & \text{Let }x=1 \\ & 3A=1,\text{ }A=\frac{1}{3} \\ & \text{Let }x=-2 \\ & -3B=1,\text{ }B=-\frac{1}{3} \\ & \frac{1}{(x-1)(x+2)}=\frac{1}{3}\left( \frac{1}{x-1}-\frac{1}{x+2} \right) \\ & \frac{1}{x-1}={{(x-1)}^{-1}}={{\left[ -(1-x) \right]}^{-1}}=-{{1}^{-1}}{{(1-x)}^{-1}} \\ & \frac{1}{x-1}=-\left[ 1-x+\frac{-1(-2)}{2}{{x}^{2}}+\frac{(-1)(-2)(-3)}{3!}{{x}^{3}}+--- \right] \\ & \frac{1}{x-1}=-\left[ 1+x+{{x}^{2}}+{{x}^{3}}+-- \right]=-1-x-{{x}^{2}}-{{x}^{3}} \\ & \frac{1}{x+2}={{(x+2)}^{-1}}={{(2+x)}^{-1}}={{2}^{-1}}{{\left( 1+\frac{x}{2} \right)}^{-1}} \\ & =\frac{1}{2}\left[ 1-\frac{x}{2}+\frac{{{x}^{2}}}{4}-\frac{{{x}^{3}}}{8}+ \right]=\frac{1}{2}-\frac{x}{4}+\frac{{{x}^{2}}}{8}-\frac{{{x}^{3}}}{16}+--- \\ & \frac{1}{(x-1)(x+2)}=\frac{1}{3}\left( \frac{1}{x-1}-\frac{1}{x+2} \right) \\ & \frac{1}{(x-1)(x+2)}=\frac{1}{3}\left[ -1-x-{{x}^{2}}-{{x}^{3}}-\left( \frac{1}{2}-\frac{x}{4}+\frac{{{x}^{2}}}{8}-\frac{{{x}^{3}}}{16}+--- \right) \right] \\ & \frac{1}{(x-1)(x+2)}=\frac{1}{3}\left[ -1-x-{{x}^{2}}-{{x}^{3}}-\frac{1}{2}+\frac{x}{4}-\frac{{{x}^{2}}}{8}+\frac{{{x}^{3}}}{16}+---- \right] \\ & \frac{1}{(x-1)(x+2)}=\frac{1}{3}\left[ \frac{3}{2}-\frac{3x}{4}-\frac{9{{x}^{2}}}{8}-\frac{15{{x}^{3}}}{16} \right] \\ & \frac{1}{(x-1)(x+2)}=\frac{1}{2}-\frac{x}{4}-\frac{3{{x}^{2}}}{8}-\frac{5{{x}^{3}}}{16} \\\end{align}$