Let's start by factorising the denominator
By trial and error we finf that #(x-2)# is a factor
Let's do a long division
#x^3+2x^2-5x-6##color(white)(aaaa)##∣##x-2#
#x^3-2x^2##color(white)(aaaaaaaaaaaa)##∣##x^2+4x+3#
#0+4x^2-5x#
#color(white)(aaa)##4x^2-8x#
#color(white)(aaaaa)##0+3x-6#
#color(white)(aaaaa)##0+3x-6#
#color(white)(aaaaaaaaa)##0-0#
So the factorisation is #x^2+4x+3=(x+1)(x+3)#
And #x^3+2x^2-5x-6=(x-2)(x+1)(x+3)#
So the partial fractiona are
#(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))#
#(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 the integration is
#int((x+13)dx)/(x^3+2x^2-5x-6)=intdx/(x-2)-int(2dx)/(x+1)+intdx/(x+3)#
#ln(x-2)-2ln(x+1)+ln(x+3)+C#
