Answer
$\displaystyle \frac{3}{x}-\frac{1}{x+1}+\frac{2}{x-5}$
Work Step by Step
(see p.541:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.)
$\displaystyle \frac{4x^{2}-5x-15}{x(x+1)(x-5)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-5}$
Multiply both sides of the equation $x(x+1)(x-5)$, the LCD.
$4x^{2}-5x-15=A(x+1)(x-5)+Bx(x-5)+Cx(x+1)$
$=A(x^{2}-4x-5)+Bx^{2}-5Bx+Cx^{2}+Cx$
$=Ax^{2}-4Ax-5A+Bx^{2}-5Bx+Cx^{2}+Cx$
$=Ax^{2}+Bx^{2}+Cx^{2}-4Ax-5Bx+Cx-5A$
$=(A+B+C)x^{2}+(-4A-5B+C)x-5A$
Equate coefficients of like powers and constant terms.
$\left\{\begin{array}{llll}
I. & -5A=-15 & \Rightarrow & A=3\\
II. & -4A-5B+C=-5 & & \\
III. & A+B+C=4 & &
\end{array}\right.$
Solve the system.
Inserting A=3 into equations II and III:
$\left\{\begin{array}{llll}
-4(3)-5B+C=-5 & \Rightarrow & -5B+C=7 & \\
3+B+C=4 & \Rightarrow & B+C=1 &
\end{array}\right.$
Subtracting the second equation from the first,
$-6B=6$
$B=-1$
Back substitute B into $B+C=1$
$-1+C=1$
$C=2$
$A=3$,
$B=-1$,
$C=2$.
$\displaystyle \frac{4x^{2}-5x-15}{x(x+1)(x-5)}=\frac{3}{x}-\frac{1}{x+1}+\frac{2}{x-5}$
