<s>[INST] 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?
 [/INST] in $x - 2 x^2 + \frac{4 x^3}{3} +C$ you did right and got a constant C.
in $-\frac{1}{6} + x - 2 x^2 + \frac{4 x^3}{3} +C$ what happened is that you got different C, you can call it D. and it is a constant so $-\frac{1}{6}$ is also a part of that same constant. your expression should be $x - 2 x^2 + \frac{4 x^3}{3} +E$ where E=D$-\frac{1}{6}$. This is a property of indefinite integrals, if you integrated on some interval then your substitution would change your boundaries of integration and you would get the same result in the end.
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