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

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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

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

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  • $\begingroup$ Yes of course! Thanks $\endgroup$
    – user
    Nov 27 '19 at 7:26
  • $\begingroup$ So $$\frac{x^2-4x-12}{2}$$ is same as $$ \frac{x^2-4x}{2}$$ ? $\endgroup$ Nov 27 '19 at 7:32
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    $\begingroup$ @JameelKhamis No, they are not the same. However, they only differ by a constant: $$\frac{x^2 - 4x - 12}{2} = \frac{x^2 - 4x}{2} + \underbrace{(-6)}_{=C}. $$ $\endgroup$
    – Xander Henderson
    Nov 28 '19 at 19:25
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Your answers differ by a constant. Therefore, if one of them is correct, the other one is correct too.

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  • $\begingroup$ so, that -12 won't make any difference to the answer, correct? $\endgroup$ Nov 27 '19 at 7:34
  • $\begingroup$ Correct. For instance, both $x$ and $x-12$ are such that their derivative is $1$. $\endgroup$ Nov 27 '19 at 7:36

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