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$\int (x+1) dx$

Substituting:

$u = x + 1$ and $du = dx$

$$\int u \,du = \frac{u^2}{2}$$

Which gives:

$$\int (x+1) \, dx = \frac{(x+1)^2}{2}$$

This is completely wrong as the correct answer is $\frac{x^2}{2}+x$

Why did this happen?

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    $\begingroup$ What's the difference of $(x+1)^2/2$ and $x^2/2+x$? $\endgroup$ Commented Sep 8, 2017 at 21:29
  • $\begingroup$ It should be $\int (x+1)dx=\frac{(x+1)^2}{2}+C$. $\endgroup$
    – user236182
    Commented Sep 8, 2017 at 21:31
  • $\begingroup$ you missed the +C $\endgroup$
    – vik1245
    Commented Sep 8, 2017 at 21:32
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    $\begingroup$ @CameronBuie It is likely a typo for $\frac{x^2}{2} + 2x$, with the wrong braces used. $\endgroup$
    – Xander Henderson
    Commented Sep 9, 2017 at 0:08
  • $\begingroup$ It is completely right, rather than wrong; a correct answer is $\frac{x^2}{2}+x$, in the sense this is an antiderivative, but $\frac{(x+1)^2}{2}$ is an antiderivative as well. $\endgroup$
    – egreg
    Commented Sep 9, 2017 at 6:03

3 Answers 3

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If you expand your second answer, you get $$ \frac{(x+1)^2}{2} = \frac{x^2}{2} + x + \frac{1}{2}. $$ This differs from the "correct" answer by a constant. But if two functions differ by a constant, they will have the same derivative. Hence both answers are equally correct (or incorrect). I submit that a truly correct answer would be $$ \frac{x^2}{2} + x + C \qquad\text{or}\qquad \frac{(x+1)^2}{2} + C, $$ where $C$ is a real constant.


Addendum: I have a stupid trick for getting my students to remember the constant. We have $$ \int \frac{1}{\text{cabin}} \, \mathrm{d}(\text{cabin}) = \text{house boat}.$$ Why?

Remember that $$ \int \frac{1}{x}\, \mathrm{d}x = \log|x| + C. $$ Thus $$ \int \frac{1}{\text{cabin}} \, \mathrm{d}(\text{cabin}) = \log|\text{cabin}| + \text{sea}, $$ which is quite clearly a house boat.

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You forgot the constant C. It is very important:

$$ \int u du = \frac{u^2}{2} + C = \frac{(x+1)^2}{2} + C = \frac{x^2}{2} + x + \frac{1}{2}+ C = \frac{x^2}{2} + x + C' $$

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Both answers are right: $\displaystyle \frac{(x+1)^2} 2 = \frac{x^2 + 2x + 1} 2 = \frac {x^2} 2 + x + \text{constant}$

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