We need to factorise the denominator
Let f(x)=x^3+2x^2-5x-6
Then
f(2)=8+8-10-6=0
Therefore,
(x-2) is a factor of f(x)
We can do a long division
color(white)(aaaa)x^3+2x^2-5x-6color(white)(aaaa)|x-2
color(white)(aaaa)x^3-2x^2color(white)(aaaaaaaaaaaa)|x^2+4x+3
color(white)(aaaaa)0+4x^2-5x
color(white)(aaaaaaa)+4x^2-8x
color(white)(aaaaaaaa)+0+3x-6
color(white)(aaaaaaaaaaaa)+3x-6
color(white)(aaaaaaaaaaaa)+0-0
so,
x^3+2x^2-5x-6=(x-2)(x^2+4x+3)
=(x-2)(x+1)(x+3)
Now, we perform the decomposition into partial fractions
(x+13)/(x^3+2x^2-5x-6)=(x+13)/((x-2)(x+1)(x+3))
=A/(x-2)+B/(x+1)+C/(x+3)
=(A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1))/((x-2)(x+1)(x+3))
The denominators are the same, we compare the numerators
x+13=A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1)
Let, x=2, =>,15=15A, =>,A=1
Let, x=-1, =>, 12=-6B, =>, B=-2
Let, x=-3, =>, 10=10C, =>, C=1
So,
(x+13)/(x^3+2x^2-5x-6)=1/(x-2)-2/(x+1)+1/(x+3)
