Using partial fractions would require lowering the degree to the numerator so that it is lower than the denominator. In this case it is easier to perform the substitution:
#u=x-1#
#du = dx#
#int (x^4dx)/(x-1)^3 = int ((u+1)^4du)/u^3#
expanding the power of the binomial:
#int (x^4dx)/(x-1)^3 = int ((u^4+4u^3+6u^2+4u+1)^4du)/u^3#
and applying linearity:
#int (x^4dx)/(x-1)^3 = int udu + 4 int du +6 int (du)/u +4 int (du)/u^2 +int (du)/u^3#
#int (x^4dx)/(x-1)^3 =u^2/2+ 4 u +6 ln abs u -4/u -1/(2u^2)+C#
#int (x^4dx)/(x-1)^3 =6 ln abs u + (u^4+ 8 u^3 -8u -1)/(2u^2)+C#
#int (x^4dx)/(x-1)^3 =6 ln abs (x-1) + ((x-1)^4+ 8 (x-1)^3 -8(x-1) -1)/(2(x-1)^2)+C#
#int (x^4dx)/(x-1)^3 =6 ln abs (x-1) + (x^4-4x^3+6x^2-4x+1 + 8x^3-24x^2+24x -8 -8 x+8-1)/(2(x-1)^2)+C#
#int (x^4dx)/(x-1)^3 =6 ln abs (x-1) + (x^4+4x^3-18x^2+12x)/(2(x-1)^2)+C#
#int (x^4dx)/(x-1)^3 =6 ln abs (x-1) + (x(x^3+4x^2-18x+12))/(2(x-1)^2)+C#
