Question

Write the partial fraction decomposition of each rational expression.

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

Short answer

The partial fraction decomposition is$\frac{-3}{4x}+\frac{-1}{2x^2}+\frac{3}{4(x-2)}$.

Step-by-step solution

Step 1. Given Information.

The rational expression is,

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

Step 2. Partial fraction decomposition.

Decomposition of the partial fraction,

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

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

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

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

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

Solving the equations we get,

$$\begin{gathered} B=-\frac{1}{2} \\ A=-\frac{3}{4} \\ C=\frac{3}{4} \end{gathered}$$

The partial fraction decomposition is,

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

Step 3. Checking the solution.

Adding the rational expressions,

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

$$\begin{gathered} =\frac{-3(x^2-2x)+(-1)(2x-4)+3x^2}{4x^2(x-2)} \\ =\frac{-3x^2+6x-2x+4+3x^2}{4x^2(x-2)} \\ =\frac{4x+4}{4x^2(x-2)} \\ =\frac{4(x+1)}{4x^2(x-2)} \\ =\frac{x+1}{x^2(x-2)} \end{gathered}$$

The solution is true.