Why is there a difference of $\frac{-1}{6}$ when integrating $(1-2x)^2$ in two different ways? I integrated $(1-2x)^2$ by expanding the expression first, that is $1 - 4 x + 4 x^2$ and I get $$x - 2 x^2 + \frac{4 x^3}{3} +C$$
When I integrate by substitution I get $$\frac{-(1-2x)^3}{6} + C$$
Expanding $(-(1-2x)^3)/6 +C$ get $$-\frac{1}{6} + x - 2 x^2 + \frac{4 x^3}{3} +C$$
Why the difference in the two answers?

Indefinite integral can differ by a constant (you denote it by $C$). In this case you just have $C=-1/6$.
This is simply given by the fact that indefinite integral (or for this purpose better name would be antiderivative) of $f(x)$ is a function (not the function) $F(x)$ such that $F'(x) = f(x)$. And since derivative of the constant is $0$ (zero), this works for both of your results, namely you have:
$$\left(x - 2 x^2 + \frac{4 x^3}{3}\right)' = 1-4x+4x^2$$
$$\left(x - 2 x^2 + \frac{4 x^3}{3}-\frac{1}{6}\right)' = 1-4x+4x^2$$
Also it is not hard to end up with another constant. For example by choosing substitution $t=2-2x$ you can easily verify that after expanding and integrating you end up with constant $C=-1/3$.
I suggest to check other answers related to this where constant is even more tricky, for example: 
Different results from the same integral with two different methods, Two different solutions to integral or Is $\int \frac{1}{2x} \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$.
