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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?

Sil
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pirsquare
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3 Answers3

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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.

Filip Skrinjar
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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 $\frac{\ln|x|}{2}+ C$.

Community
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Sil
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When you integrate a indefinite integral, the function you get is not unique. That's why we add the constant $C$ to represent all those functions.

The equation $y=\int 2dx$ can yield $y=2x$ or $y=2x+1$, or $y=2x+50$, it doesn't matter. We know for sure that $y=2x+(some\,constant\,number)$ since the constant would be zero anyway if we get the derivative with respect to $x$. So in your example, it would be more appropriate to label the constants with different subscripts, $C_0$ and $C_1$ for example, since they can be unequal.

Euhi
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