#### Answer

$\dfrac{5}{x+2}+\dfrac{3}{x+1}+\dfrac{2}{x-1}$

#### Work Step by Step

Here, we have $\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}=\dfrac{A}{x+2}+\dfrac{B}{x+1}+\dfrac{C}{x-1}$
$10x^2+9x-7 =A(x-1)(x+1)+B(x+2)(x-1)+C(x+2)(x+1)$
or, $10x^2+9x-7 =(A+B+C)x^2+(B+3C)x-A-2B+2C$
Equating the coefficients, we get
$A+B+C=10$ and $B+3C =9 \implies B=-2$
and $-A-2B+2C=-7$
$ \implies (A+B+C)+(-A-2B+2C)=10-7$ gives: $C=2$
and $ B+3(2)=9 \implies B=3$
$A+3+2= 10 \implies A=5$
Hence, $\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}=\dfrac{5}{x+2}+\dfrac{3}{x+1}+\dfrac{2}{x-1}$