91Ó°ÊÓ

The rational expression is,

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

Decomposition of the rational expression,

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

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

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

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

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

Solving equation (3) we get,$A=1-B$.

Inputting the value of A in equations (2) and (3), we get,

$$\begin{gathered} 3B+C=3 \\ -3B+2C=-1 \end{gathered}$$

Solving the equations, we get,

$$\begin{gathered} C=\frac{2}{3} \\ B=\frac{7}{9} \\ A=\frac{2}{9} \end{gathered}$$

The partial fraction decomposition is,

$\frac{x^2+x}{(x+2){(x-1)}^2}=\frac{{\displaystyle \frac{2}{9}}}{(x+2)}+\frac{{\displaystyle \frac{7}{9}}}{x-1}+\frac{{\displaystyle \frac{2}{3}}}{{(x-1)}^2}$

$=\frac{2}{9(x+2)}+\frac{7}{9(x-1)}+\frac{2}{3{(x-1)}^2}$.

Adding the rational expressions,

$\frac{2}{9(x+2)}+\frac{7}{9(x-1)}+\frac{2}{3{(x-1)}^2}=\frac{2{(x-1)}^2+7(x-1)(x+2)+2\left(3\right(x+2)}{9(x+2){(x-1)}^2}$

$$\begin{gathered} =\frac{2(x^2+1-2x)+7(x^2+x-2)+6(x+2)}{9(x+2){(x-1)}^2} \\ =\frac{2x^2+2-4x+7x^2+7x-14+6x+12}{9(x+2){(x-1)}^2} \\ =\frac{9x^2+9x}{9(x+2){(x-1)}^2} \\ =\frac{9(x^2+x)}{9(x+2){(x-1)}^2} \\ =\frac{x^2+x}{(x+2){(x-1)}^2} \end{gathered}$$

The solution is true.