Answer
$ \displaystyle \frac{3}{x-2}+\frac{2}{x+1}$
Work Step by Step
Both factors of the denominator are distinct linear factors...
see p.541. Decomposition with Distinct Linear Factors In the Denominator:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.
Setup:
$\displaystyle \frac{5x-1}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\qquad $
multiply both sides with $(x-2)(x+1)\quad$(LCD)
$5x-1=A(x+1)+B(x-2)$
$ 5x-1=Ax+A+Bx-2B\qquad$ /group like terms...
$5x-1=x(A+B)+(A-2B)$
... we now equate the coefficients of like powers...
$\left\{\begin{array}{l}
5=A+B\\
-1=A-2B
\end{array}\right.\qquad $solve the system
Subtract the second equation from the first:
$5+1=A-A+B+2B$
$6=3B$
$B=2$
back substitute:$5=A+B$
$5=A+2$
$A=3$
Rewrite the setup decomposition with $A=3, B=2$
$\displaystyle \frac{5x-1}{(x-2)(x+1)}= \frac{3}{x-2}+\frac{2}{x+1}$
