Answer
$ \displaystyle \frac{3}{x}+\frac{2}{x-1}-\frac{1}{x+3}$
Work Step by Step
(see p.541:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.)
$\displaystyle \frac{4x^{2}+13x-9}{x(x-1)(x+3)}=\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x+3)}$
multiply both sides with $x(x-1)(x+3)\quad $(LCD)
$4x^{2}+13x-9=$
$A(x-1)(x+3)+Bx(x+3)+Cx(x-1)$
$4x^{2}+13x-9=$
$A(x^{2}+2x-3)+B(x^{2}+3x)= C(x^{2}-x)$
$4x^{2}+13x-9=$
$x^{2}(A+B+C)+x(2A+3B-C)+(-3A)$
Constant terms: $-3A=-9\Rightarrow A=3 $
equate the coefficients of like powers, insert A=$-3$...
$\left\{\begin{array}{lll}
I. & 4=3+B+C & \\
II. & 13=2(3)+3B-C &
\end{array}\right.$
$\left\{\begin{array}{lll}
I. & 1=B+C & \\
II. & 7=3B-C &
\end{array}\right.$
$(II+I)$ eliminates C ...
$8=4B$
$B=2,$ back-substitute into $1=B+C$
$1=2+C$
$C=-1$
Rewrite the decomposition with $A=3, B=2, C=-1$
$\displaystyle \frac{4x^{2}+13x-9}{x(x-1)(x+3)}= \frac{3}{x}+\frac{2}{(x-1)}-\frac{1}{(x+3)}$
