Note that the denominator is:
#x^2+2x+1 = (x+1)^2#
so we can substitute #t=x+1#, #x=t-1#, #dx=dt# and have:
#int (x^3dx)/(x^2+2x+1) = int ((t-1)^3dt)/t^2#
#int (x^3dx)/(x^2+2x+1) = int ((t^3-3t^2+3t-1)dt)/t^2#
#int (x^3dx)/(x^2+2x+1) = int (t-3+3/t-1/t^2)dt#
Using the linearity of the integral:
#int (x^3dx)/(x^2+2x+1) = int tdt -3int dt +3int (dt)/t-int (dt)/t^2#
#int (x^3dx)/(x^2+2x+1) = t^2/2 -3t +3lnabst+1/t+C#
and undoing the substitution:
#int (x^3dx)/(x^2+2x+1) = (x+1)^2/2 -3(x+1) + 1/(x+1) + 3lnabs(x+1)+C#
Simplifying:
#int (x^3dx)/(x^2+2x+1) = ((x+1)^3-6(x+1)^2 +2)/(2(x+1)) + 3lnabs(x+1)+C#
#int (x^3dx)/(x^2+2x+1) = (x^3-3x^2-9x-3)/(2(x+1)) + 3lnabs(x+1)+C#
