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If $f(x)=x-1$, I want to find $$ \int (x-1) \ dx $$. I make a variable substitution z = x-1, dx = dz so it becomes $$ \int z \ dz \\ = \frac{z^2}{2} + C $$. If I now substitute back $z=x-1,$ I get: $$ \int x-1 \ dx = \frac{(x-1)^2}{2} + C $$

Now if I do it another way without variable substitution I get: $$ \int x \ dx - \int 1 \ dx = \frac{x^2}{2} - x + C$$

What is going on here since these answers seem different to me? Are they both right but the constans $C$ are different or what am I missing here?

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    $\begingroup$ Two antiderivatives differs in a constant. Just expand the 1st solution $\endgroup$ Commented Aug 19, 2021 at 1:13
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    $\begingroup$ This is why I hate indefinite integrals and the “$+C$”. $\endgroup$ Commented Aug 19, 2021 at 1:39
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    $\begingroup$ One way to avoid such ambiguity is to work with definite integrals instead, e.g., $$F_0(x)=\int_0^x (x'-1)\,dx'=\frac12 x^2-x,\quad F_1(x) =\int_1^x (x'-1)\,dx'=\frac12 (x-1)^2.$$ Note that $F_1(x)-F_0(x)=\int_0^1 (x-1)\,dx = -1/2$, so the two indeed differ by a constant. Whether this is an improvement will, in part, depend on your stomach for dummy variables. $\endgroup$ Commented Aug 19, 2021 at 1:42
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    $\begingroup$ @Deane I absolutely agree. Every time I see a $\int$ without a domain of integration I cringe - what does it mean geometrically? If one wants an arbitrary integral they should write $\int_a^x (t-1)\,\textrm{d}t$ for an arbitrary constant $a$. $\endgroup$ Commented Aug 19, 2021 at 1:53
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    $\begingroup$ Check the derivatives of both the answers, i.e, another way to check if they differ by a constant. $\endgroup$ Commented Aug 19, 2021 at 3:20

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Both answers are correct. Indefinite integrals are defined up to a constant C.

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Both answers are correct.

$$\begin{align}\frac{(x-1)^2}{2}&=\frac{x^2-2x+1}{2}\\ &=\frac{x^2}2-x+\frac{1}{2} \end{align}$$

So your two solutions differ by a constant.

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  • $\begingroup$ Which equivalent answer is considered the most simplified or generally preferred form? $\endgroup$ Commented Aug 19, 2021 at 1:30
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    $\begingroup$ There isn’t really a preferred answer, as far as I know. I think $x^2/2-x+C$ solution is simpler to derive, and generalizes to all polynomials, but there is an elegance to the substitution method, too, and might even be preferred when integrating something like $(x-1)^3.$ $\endgroup$ Commented Aug 19, 2021 at 1:34
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First solution: $${\dfrac {(x-1)^2}{2} + C = \dfrac {(x^2-2x+1)}{2}+C = \dfrac {x^2}{2}-x + \dfrac {1}{2}+C}.$$ The $\dfrac {1}{2} +C$ is also constant; thus, both solutions are correct.

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  • $\begingroup$ Which equivalent answer (entailing both solutions differing by a constant of integration) is considered the most simplified or generally preferred form? The most expanded vs factorized one? $\endgroup$ Commented Aug 19, 2021 at 1:31
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    $\begingroup$ Why write $\rightarrow$ when $=$ will do? $\endgroup$ Commented Aug 19, 2021 at 1:35
  • $\begingroup$ It depends on how much you want it simplified. If you prefer your solution neater, use the first method of substitution and do not expand your solution; otherwise, you can integrate term separately. $\endgroup$ Commented Aug 19, 2021 at 1:36
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    $\begingroup$ @ThomasAndrews You know, I had that first and then thought arrows would be better, but you're right - equals signs work much better. I've corrected my post. $\endgroup$ Commented Aug 19, 2021 at 1:37

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