Problem with integration by substitution. 
Using the substitution $u=1-x$, compute the integral of
$\int{x(1-x)^2}dx.$

My Work: Let $u=1-x.$ Then $\mathrm dx=-\mathrm du$ and $x=1-u,$ so
$$\int{x(1-x)^2}\,\mathrm dx\\
=-\int{(1-u)u^2}\,\mathrm du\\
=\frac{u^4}{4}-\frac{u^3}{3}+C\\
=\frac{(1-x)^4}{4}-\frac{(1-x)^3}{3}+C\\
=\frac{3x^4-8x^3+6x^2-1}{12}+C.$$
However, the answer gotten by expanding the brackets first then integrating does not seem to match that of the answer above arrived by substitution:
$$\int{x(1-x)^2}\,\mathrm dx\\
=\int{x-2x^2+x^3}\,\mathrm dx\\
=\frac{x^2}{2}-\frac{2x^3}{3}+\frac{x^4}{4}+C.$$
I cannot account for the $-\frac{1}{12}$ from the method using substitution.

Let the domain of $f$ be an interval.
Then every pair of the infinitely many antiderivatives of $f$ differ by a constant. Since the indefinite integral $$\int f(x)\,\mathrm dx$$ gives the general specification of these antiderivatives, it has the form $F(x)+C,$ where the constant of integration $C$ is arbitrary.
Both your answers are correct, and their specified constants of integration differ by $C_2-C_1=\frac1{12}.$ To simplify the first answer, substitute $$C:=D+\frac1{12}$$ so that $$\int{x(1-x)^2}\,\mathrm dx=\frac{3x^4-8x^3+6x^2}{12}+D.$$
