Write the partial fraction decomposition of each rational expression.

$\frac{x-3}{(x+2){(x+1)}^2}$.

The rational expression is,

$\frac{x-3}{(x+2){(x+1)}^2}$.

Decomposition of rational expressions,

$\frac{x-3}{(x+2){(x+1)}^2}=\frac{A}{x+2}+\frac{B}{x+1}+\frac{C}{{(x+1)}^2}...........\left(1\right)$

Multiplying both sides $(x+2){(x+1)}^2$,

role="math" localid="1647507361646" $$\begin{gathered} x-3=A{(x+1)}^2+B(x+1)(x+2)+C(x+2) \\ x-3=A(x^2+2x+1)+B(x^2+3x+2)+C(x+2) \\ x-3=A{x}^2+2Ax+A+B{x}^2+3Bx+2B+Cx+2C \\ x-3=(A+B)x^2+(2A+3B+C)x+(A+2B+2C).........\left(2\right) \end{gathered}$$

Equating the coefficients of the like powers of $x$, we get,

role="math" localid="1647507371191" $$\begin{gathered} A+B=0......\left(3\right) \\ 2A+3B+C=1......\left(4\right) \\ A+2B+2C=-3.......\left(5\right) \end{gathered}$$

Solving equation $\left(3\right)$, we get, $A=-B$.

Inputting the value of A in both equations (4) and (5) we get,

role="math" localid="1647507389528" $$\begin{gathered} B+C=1 \\ B+2C=-3 \end{gathered}$$

Solving we get the values as,

role="math" localid="1647507430568" $$\begin{gathered} C=-4 \\ B=5 \\ A=-5 \end{gathered}$$

The partial fraction decomposition is,

$\frac{x-3}{(x+2){(x+1)}^2}=\frac{{\displaystyle -5}}{(x+2)}+\frac{{\displaystyle 5}}{x+1}+\frac{{\displaystyle -4}}{{(x+1)}^2}$.

Adding the rational expressions,

$\frac{-5}{x+2}+\frac{5}{x+1}+\frac{-4}{{(x+1)}^2}=\frac{-5{(x+1)}^2+5(x+2)(x+1)+(-4)(x+2)}{(x+2){(x+1)}^2}$

$$\begin{gathered} =\frac{-5(x^2+2x+1)+5(x^2+3x+2)+(-4)(x+2)}{(x+2){(x+1)}^2} \\ =\frac{-5x^2-10x-5+5x^2+15x+10-4x-8}{(x+2){(x+1)}^2} \\ =\frac{x-3}{(x+2){(x+1)}^2} \end{gathered}$$

The solution is true.