Why integration by long division gives different answer than using u substitution?

Question

So, after solving this question, I got two different answers? and I don't think it's supposed to be this way?

Evaluate the integral $$\int \frac{x^2+2}{x+2} dx$$

Using polynomial long division, I get $$\frac{x^2}2-2x+6\ln|x+2|+C,$$ but using substitution, I get $$\frac{(x+2)^2}2-4(x+2)+6\ln|x+2|+C.$$

The second solution is the same as the first, up to a difference in $C$. — , Nov 27 '19 at 07:12
Possible duplicate: https://math.stackexchange.com/questions/2816513/same-integration-with-2-different-answers — Xander Henderson, Nov 27 '19 at 13:56
https://math.stackexchange.com/questions/3453558/getting-different-answers-when-integrating-using-different-techniques — Xander Henderson, Nov 27 '19 at 18:58
They differ only by an arbitrary constant. First one is $x^2-2x+6, $ and the second one is $x^2-2x-6 $ — Narasimham, Nov 27 '19 at 19:58

user · Answer 1 · 2019-11-27T13:32:05.403

1

Note that

$$\frac{(x+2)^2}2-4(x+2)+6\ln|x+2|+C=\frac{x^2}2-2x-6+6\ln|x+2|+C=$$

$$=\frac{x^2}2-2x+6\ln|x+2|+(C-6)=\frac{x^2}2-2x+6\ln|x+2|+C_1$$

therefore the two results are the same up to a constant which is not essential, indeed in both cases

$$\frac{d}{dx}\left(\frac{x^2}2-2x+6\ln|x+2|+C\right)=\frac{x^2+2}{x+2}$$

edited Nov 27 '19 at 13:32

answered Nov 27 '19 at 07:13

user

Yes of course! Thanks – user Nov 27 '19 at 07:26
So $$\frac{x^2-4x-12}{2}$$ is same as $$ \frac{x^2-4x}{2}$$ ? – Jameel Khamis Nov 27 '19 at 07:32
1

@JameelKhamis No, they are not the same. However, they only differ by a constant: $$\frac{x^2 - 4x - 12}{2} = \frac{x^2 - 4x}{2} + \underbrace{(-6)}_{=C}. $$ – Xander Henderson Nov 28 '19 at 19:25

score -1 · Answer 2 · answered Nov 27 '19 at 07:14

-1

Your answers differ by a constant. Therefore, if one of them is correct, the other one is correct too.

answered Nov 27 '19 at 07:14

so, that -12 won't make any difference to the answer, correct? – Jameel Khamis Nov 27 '19 at 07:34
Correct. For instance, both $x$ and $x-12$ are such that their derivative is $1$. – José Carlos Santos Nov 27 '19 at 07:36

2 Answers2