Answer
The partial fraction is $\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$.
Work Step by Step
Let us consider the decomposition as,
$\begin{align}
& \frac{7x-4}{{{x}^{2}}-x-12}=\frac{7x-4}{\left( x+3 \right)\left( x-4 \right)} \\
& =\frac{A}{\left( x+3 \right)}+\frac{B}{\left( x-4 \right)}
\end{align}$
Then, multiply $\left( x+3 \right)\left( x-4 \right)$ on both sides:
$\begin{align}
& \left( x+3 \right)\left( x-4 \right)\frac{7x-4}{\left( x+3 \right)\left( x-4 \right)}=\left( x+3 \right)\left( x-4 \right)\frac{A}{\left( x+3 \right)}+\left( x+3 \right)\left( x-4 \right)\frac{B}{\left( x-4 \right)} \\
& 7x-4=\left( x-4 \right)A+\left( x+3 \right)B \\
& =Ax-4A+Bx+3B \\
& 7x-4=x\left( A+B \right)-4A+3B
\end{align}$ …… (I)
Now, compare the coefficient of equation (I):
$A+B=7$ …… (II)
$-4A+3B=-4$ …… (III)
Then, multiply equation (II) by 4 and add with equation (III):
$\begin{align}
& 7B=24 \\
& B=\frac{24}{7}
\end{align}$
Now, put the value of B in equation (II):
$\begin{align}
& A+\frac{24}{7}=7 \\
& A=7-\frac{24}{7} \\
& =\frac{49-24}{7} \\
& =\frac{25}{7}
\end{align}$
Therefore,
$\frac{7x-4}{{{x}^{2}}-x-12}=\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$
Thus, the partial fraction of the given expression is $\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$.
