We need to factorise the denominator
Let f(x)=x^3+2x^2-5x-6f(x)=x3+2x2−5x−6
Then
f(2)=8+8-10-6=0f(2)=8+8−10−6=0
Therefore,
(x-2)(x−2) is a factor of f(x)f(x)
We can do a long division
color(white)(aaaa)aaaax^3+2x^2-5x-6x3+2x2−5x−6color(white)(aaaa)aaaa|∣x-2x−2
color(white)(aaaa)aaaax^3-2x^2x3−2x2color(white)(aaaaaaaaaaaa)aaaaaaaaaaaa|∣x^2+4x+3x2+4x+3
color(white)(aaaaa)aaaaa0+4x^2-5x0+4x2−5x
color(white)(aaaaaaa)aaaaaaa+4x^2-8x+4x2−8x
color(white)(aaaaaaaa)aaaaaaaa+0+3x-6+0+3x−6
color(white)(aaaaaaaaaaaa)aaaaaaaaaaaa+3x-6+3x−6
color(white)(aaaaaaaaaaaa)aaaaaaaaaaaa+0-0+0−0
so,
x^3+2x^2-5x-6=(x-2)(x^2+4x+3)x3+2x2−5x−6=(x−2)(x2+4x+3)
=(x-2)(x+1)(x+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))x+13x3+2x2−5x−6=x+13(x−2)(x+1)(x+3)
=A/(x-2)+B/(x+1)+C/(x+3)=Ax−2+Bx+1+Cx+3
=(A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1))/((x-2)(x+1)(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)x+13=A(x+1)(x+3)+B(x−2)(x+3)+C(x−2)(x+1)
Let, x=2x=2, =>⇒,15=15A15=15A, =>⇒,A=1A=1
Let, x=-1x=−1, =>⇒, 12=-6B12=−6B, =>⇒, B=-2B=−2
Let, x=-3x=−3, =>⇒, 10=10C10=10C, =>⇒, C=1C=1
So,
(x+13)/(x^3+2x^2-5x-6)=1/(x-2)-2/(x+1)+1/(x+3)x+13x3+2x2−5x−6=1x−2−2x+1+1x+3