Answer
$\dfrac{4x^2-7x-3}{x(x-1)(x+1)}=\dfrac{3}{x}-\dfrac{3}{x-1}+\dfrac{4}{x+1}$
Work Step by Step
We are given the fraction:
$\dfrac{4x^2-7x-3}{x(x-1)(x+1)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{4x^2-7x-3}{x(x-1)(x+1)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}$
Multiply all terms by the least common denominator $x(x-1)(x+1)$:
$x(x-1)(x+1)\cdot\dfrac{4x^2-7x-3}{x(x-1)(x+1)}=x(x-1)(x+1)\cdot\dfrac{A}{x}+x(x-1)(x+1)\cdot\dfrac{B}{x-1}+x(x-1)(x+1)\cdot\dfrac{C}{x+1}$
$4x^2-7x-3=A(x-1)(x+1)+Bx(x+1)+Cx(x-1)$
$4x^2-7x-3=Ax^2-A+Bx^2+Bx+Cx^2-Cx$
$4x^2-7x-3=(A+B+C)x^2+(B-C)x-A$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B+C=4\\
B-C=-7\\
-A=-3
\end{cases}$
Solve the system:
$A=3$
$\begin{cases}
3+B+C=4\\
B-C=-7
\end{cases}$
$\begin{cases}
B+C=1\\
B-C=-7
\end{cases}$
$B+C+B-C=1-7$
$2B=-6$
$B=-3$
$B+C=1$
$-3+C=1$
$C=4$
The partial fraction decomposition is:
$\dfrac{4x^2-7x-3}{x(x-1)(x+1)}=\dfrac{3}{x}-\dfrac{3}{x-1}+\dfrac{4}{x+1}$
