Answer
The partial fraction is, $\frac{3}{\left( x-2 \right)}+\frac{2}{\left( x+1 \right)}$
Work Step by Step
We have to apply the partial fraction decomposition as given below,
$\frac{5x-1}{\left( x-2 \right)\left( x+1 \right)}=\frac{A}{\left( x-2 \right)}+\frac{B}{\left( x+1 \right)}$
Now, multiply $\left( x-2 \right)\left( x+1 \right)$ on both sides,
$\begin{align}
& \left( x-2 \right)\left( x+1 \right)\frac{5x-1}{\left( x-2 \right)\left( x+1 \right)}=\left( x-2 \right)\left( x+1 \right)\frac{A}{\left( x-2 \right)}+\left( x-2 \right)\left( x+1 \right)\frac{B}{\left( x+1 \right)} \\
& 5x-1=\left( x+1 \right)A+\left( x-2 \right)B \\
& =Ax+A+Bx-2B \\
& 5x-1=x\left( A+B \right)+A-2B
\end{align}$ …… (I)
Then, compare the coefficient of equation (I),
$A+B=5$ …… (II)
$A-2B=-1$ …… (III)
Then, subtract equation (II) from equation (III),
Then,
$-3B=-6$
$\begin{align}
& B=\frac{-6}{-3} \\
& =2
\end{align}$
Now, put the value in equation (II),
$\begin{align}
& A+2=5 \\
& A=5-2 \\
& =3
\end{align}$
Therefore,
$\frac{5x-1}{\left( x-2 \right)\left( x+1 \right)}=\frac{3}{\left( x-2 \right)}+\frac{2}{\left( x+1 \right)}$
Thus, the partial fraction of the given expression is $\frac{3}{\left( x-2 \right)}+\frac{2}{\left( x+1 \right)}$.
