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.

Well
\begin{align}
\int{x(1-x)^2}dx &= \int x^3-2x^2+x \mathrm{d}x \\
&=\frac{1}{4}x^4-\frac{2}{3}x^3+\frac{x^2}{2} +C_1
\end{align}
Whereas with $u=1-x \, \implies \mathrm{d}u=-\mathrm{d}x$ and
\begin{align}
\int (1+u)^2u (- \mathrm{d}u) &= -\int u^3+2u^2+u \mathrm{d}u \\
&=-\frac{u^4}{4}-\frac{2}{3}u^3-\frac{u^2}{2}+C_2
\end{align}
Now substitute in $x=1-u$ to see equivalence.
