Answer
$[\frac{1}{5}][\frac{4}{x-3} + \frac{1}{x+2}]$
Work Step by Step
We must first find the Partial Fraction Decomposition:
$\frac{A}{x-3} + \frac{B}{x+2} = \frac{x+1}{(x-3)(x+2)}$
We must then solve for the constants:
$A(x+2) + B(x-3) = x+1$
$Ax + 2A + Bx - 3B = x +1$
$Ax + Bx = x$
$2A - 3B = 1$
This can be represented with the following matrix:
$\begin{bmatrix}
1 & 1 & |1\\
2 & -3 & |1\\
\end{bmatrix}$ ~ $\begin{bmatrix}
1 & 1 & |1\\
0 & -5 & |-1\\
\end{bmatrix}$
$-5B = -1$
$B = \frac{1}{5}$
$A + B = 1$
$A + \frac{1}{5} = 1$
$A = \frac{4}{5}$
The Partial Fraction is:
$[\frac{1}{5}][\frac{4}{x-3} + \frac{1}{x+2}]$
Checking the Result:
$[\frac{1}{5}]\frac{4x+8+x-3}{(x-3)(x+2)} = [\frac{1}{5}]\frac{5x + 5}{(x-3)(x+2)} = \frac{x+1}{(x-3)(x+2)}$
Therefore, the answer is correct.
