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I'm trying to compute the following integral $$\int \frac{7x}{(2x+1)} dx$$ Unfortunately Wolfram Alpha gives me a different result, but other integral calculators say that my result is correct. So where is my error:

$$\int \frac{7x}{(2x+1)} dx = \frac{7}{2}\int \frac{2x}{(2x+1)} dx$$ $$\frac{7}{2}\int \frac{2x+1}{(2x+1)} - \frac{1}{(2x+1)} dx = \frac{7}{2}\int 1 dx - \int \frac{1}{(2x+1)} dx$$ Let $u=2x+1$ $$=\frac{7}{2}(x-\frac{1}{2}\int\frac{1}{u}du)=\frac{7x}{2}-\frac{7}{4}\ln(|2x+1|)+C$$ Wolfram Alpha tells me that it is $$\frac{7}{4} (2 x - \ln(2 x + 1) + \underbrace{1}_{different}) + C$$ Why is there an additional $1$?

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    $\begingroup$ It doesn't matter since this is a constant term which can be "hidden" or "absorbed" by the integration constant $C$. It might happen that sometime constants pop up when computing integral with different method but in the end it doesn't matter. $\endgroup$
    – Zubzub
    Commented Mar 13, 2017 at 21:15
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    $\begingroup$ As general advice for indefinite integration problems, the answer should be "checked" by differentiation. Thinking about it that way makes it clear why "an additional $1$" in the result does not affect the result's correctness. $\endgroup$
    – hardmath
    Commented Mar 13, 2017 at 21:26
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    $\begingroup$ Just out of curiosity, why was this voted off-topic? I think it is a reasonably asked question. $\endgroup$ Commented Mar 13, 2017 at 21:30
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    $\begingroup$ Now a good question would be why WolframAlpha has the +1 when it already has +C. $\endgroup$ Commented Mar 13, 2017 at 21:37

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Note that $C$ is an arbitrary constant, so both answers are actually correct. One may thus let $k=\frac{7}{4}+C$ and then obtain the same answer you've obtained.

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    $\begingroup$ Damn you had an answer before I even had time to click "1 new question available". +1 $\endgroup$
    – mrnovice
    Commented Mar 13, 2017 at 21:15
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    $\begingroup$ @mrnovice There was a blog post about these users. We call them 'gunslingers' ;) $\endgroup$ Commented Mar 13, 2017 at 21:18
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    $\begingroup$ That is some way too fast "projectile motion" for my typing speed :-) $\endgroup$
    – user409521
    Commented Mar 13, 2017 at 21:19
  • $\begingroup$ Thank you for the answer. But what if the integral would be indefinite. Would I still obtain the same result the way I did it, when comparing to Wolfram Alpha (implying there was no error in my computation)? $\endgroup$
    – WaldoRozir
    Commented Mar 14, 2017 at 13:14
  • $\begingroup$ The integral you've mentioned is indefinite. $\endgroup$ Commented Mar 14, 2017 at 17:43
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If you look at what Wolfram gives you

$$\frac{7}{4} (2 x - \ln(2 x + 1) +1 )+ C$$

and see that you have an "extra" term, look at it this way: any constant terms that differ from your result may have been "absorbed" into the arbitrary constant.

What you get is an antiderivative, the important thing when you evaluate indefinite integrals is that you can get back to the original integrand when you take the derivative of your result.

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If $y=F(x)$ be a primitive of the function $y=f(x)$ then for every real $k$ the function $y=F(x)+k$ is a primitive also.

The primitive of $\dfrac{7x}{2x+1}$ is $\dfrac{7}{4} (2x -\ln(2 x + 1))$, then for every real $k$ the function $y=\dfrac{7}{4} (2x -\ln(2 x + 1))+k$ is a primitive also.

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