Write the partial fraction decomposition of each rational expression.

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

The rational expression is,

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

Decomposition of Rational Expressions,

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

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

$$\begin{gathered} x+1=Ax{(x-2)}^2+B{(x-2)}^2+C{x}^2(x-2)+D{x}^2 \\ x+1=Ax(x^2-4x+4)+B(x^2-4x+4)+C{x}^2(x-2)+D{x}^2 \\ x+1=A{x}^3-4A{x}^2+4Ax+B{x}^2-4Bx+4B+C{x}^3-2C{x}^2+D{x}^2 \\ x+1=(A+C)x^3+(-4A+B-2C+D)x^2+(4A-4B)x+4B..............\left(2\right) \end{gathered}$$

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

$$\begin{gathered} A+C=0..............\left(3\right) \\ -4A+B-2C+D=0...........\left(4\right) \\ 4A-4B=1...............\left(5\right) \\ 4B=1...............\left(6\right) \end{gathered}$$

Solving the equation (6), we get, $B=\frac{1}{4}$

Solving the equation (5), we get, $A=\frac{1}{2}$

Solving the equation (3), we get, role="math" localid="1647509291235" $C=-\frac{1}{2}$

Solving the equation (4), we get,$D=\frac{3}{4}$

The partial fraction decomposition is,

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

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

Adding the rational expressions,

$\frac{1}{2x}+\frac{1}{4x^2}+\frac{-1}{2(x-2)}+\frac{3}{4{(x-2)}^2}=\frac{2x{(x-2)}^2+{(x-2)}^2+(-1)\left(2x^2\right(x-2\left)\right)+3x^2}{4x^2{(x-2)}^2}$

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

The solution is true.