Perform a long division first
#(4x^3)/(x^3+2x^2-x-2)=4+(-8x^2+4x+8)/(x^3+2x^2-x-2)#
Factorise the denominator
#x^3+2x^2-x-2=(x-1)(x^2+3x+2)=(x-1)(x+1)(x+2)#
Perform the decomposition into partial fractions
#(-8x^2+4x+8)/(x^3+2x^2-x-2)=A/(x-1)+B/(x+1)+C/(x+2)#
#=(A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1))/(x^3+2x^2-x-2)#
The denominators are the same, compare the numerators
#-8x^2+4x+8=A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)#
Let #x=1#, #=>#, #4=6A#, #=>#, #A=2/3#
Let #x=-1#, #=>#, #-4=-2B#, #=>#, #B=2#
Let #x=-2#, #=>#, #-32=3C#, #=>#, #C=-32/3#
Therefore,
#(4x^3)/(x^3+2x^2-x-2)=4+(2/3)/(x-1)+(2)/(x+1)+(-32/3)/(x+2)#
And finally perform the integration,
#int((4x^3)dx)/(x^3+2x^2-x-2)=int4dx+2/3int(dx)/(x-1)+2int(dx)/(x+1)-32/3int(dx)/(x+2)#
#=4x+2/3ln(|x-1|)+2ln(|x+1|)-32/3ln(|x+2|)+C#
