Answer
$[\frac{1}{6}][\frac{-1}{2x+3} + \frac{1}{2x-3}]$
Work Step by Step
We must first find the Partial Fraction Decomposition:
$\frac{A}{2x+3} + \frac{B}{2x-3} = \frac{1}{(2x+3)(2x-3)}$
We must then solve for the constants:
$A(2x-3) + B(2x+3) = 1$
$2Ax - 3A + 2Bx + 3B = 1$
$2Ax + 2Bx = 0$
$-3A + 3B = 1$
This can be represented with the following matrix:
$\begin{bmatrix}
2 & 2 & |0\\
-3 & 3 & |1\\
\end{bmatrix}$ ~ $\begin{bmatrix}
1 & 1 & |0\\
-3 & 3 & |1\\
\end{bmatrix}$ ~ $\begin{bmatrix}
1 & 1 & |0\\
0 & 6 & |1\\
\end{bmatrix}$
$6B = 1$
$B = \frac{1}{6}$
$A + \frac{1}{6} = 0$
$A = \frac{-1}{6}$
The Partial Fraction is:
$[\frac{1}{6}][\frac{-1}{2x+3} + \frac{1}{2x-3}]$
Checking the Result:
$[\frac{1}{6}]\frac{-2x+3+2x+3}{(2x+3)(2x-3)} = [\frac{1}{6}]\frac{6}{(2x+3)(2x-3)} = \frac{1}{(2x+3)(2x-3)}$
Therefore, the answer is correct.
