This is my first post here and I searched the internet for things like "can u substitution yield an invalid result when finding an antiderivative?" and "manual integration and maxima produce different results" but this didn't find any useful links, which is why I'm posting here.


I have a rather simple expression of which I want to find the indefinite integral/antiderivative of:

$$\frac{1}{y - \frac{115}{3}}$$

For context, this is part of a problem, where I am asked to solve the following ODE without using a SolveODE command:

$$\frac{dy}{dt} = -\frac{3}{50} \; \left(y - \frac{115}{3} \right)$$

Looking at the solution provided with the problem, I see that integrating the expression described at the beginning of this question should result in the following:

$$ \ln \left( \left|x - \frac{115}{3}\right| \right) + C $$

However, when solving using GeoGebra CAS to calculate the Integral (my exact input was Integral(1 / (x - 115 / 3))), I get the following output:

$$\ln \left( \left|3 \; x - 115\right| \right) + C$$

Using external help

Because I think that the antiderivative included in the official solution does not equal the one calculated by the Integral command in GeoGebra, I went to Integral-Calculator.com (the link leads to my calculation on the website) to find out what I did wrong.

I typed in my expression (1 / (y - 115 / 3)) and told it to solve for $y$.

A surprising result

Integral-Calcularot.com found both of the solutions.
Its results section has two parts: One describing a "manual" solve (whatever that means) and one which displays the solution calculated by the Maxima software:

See screenshot

Several questions arise

  • How can there be two different (and conflicting) solutions to one math problem?
  • What did I miss?
  • Is one of the two solutions invalid?
  • Are the two solutions actually equal?


AFTER typing my entire question, this website provided me with the following other (simular) question:

Different results when integrating 1/(x^2-9) with computer tools

I'm not sure if this would be considered a "duplicate" of the other one, but it doesn't help me understand my issue better; i.e. the two answers (or that question in and of itself) doesn't completely cover what I'm trying to do.

Thanks for reading!


I've been testing and fiddling around a bit more and noticed the following:

This is what happened

It turns out flipping the fraction on its head and using $^{-1}$ leads to the desired behaviour. Any idea why?

  • 1
    $\begingroup$ Consider that C is an unspecified constant and that $ln(ab)=ln(a)+ln(b).$ $\endgroup$
    – Paul
    May 15, 2020 at 14:46
  • 1
    $\begingroup$ The "solution" of an indefinite integral is a set, that is, the expression $f(x)+C$ for arbitrary constant represent a set of solutions. The solution can be represented in various ways but the resulting set solution must be the same $\endgroup$
    – Masacroso
    May 15, 2020 at 15:46
  • $\begingroup$ Does this answer your question? Getting different answers when integrating using different techniques $\endgroup$
    – Sil
    May 15, 2020 at 18:14

1 Answer 1


(this is my own answer)

My personal conclusion

Some questions are still open! (see below)

I tried around a bit more (see the "Update" section of my(=this) question), and it appears to be a quirk of GeoGebra.

Here is another example:


This behavior only occurs when the expressions are inserted directly into the Integral(...) command. If they are row-references (eg. Integral($1)), or defined constants (first, write A:=... and then in another line Integral(A)) this doesn't happen, and both will result in the expression calculated in line 4.

Confusing findings

This question is now closed, but only in a practical sense. As in: In practice, GeoGebra threats the inputs differently.

However, I still don't know if GeoGebra calculated correctly both times.

If we consider this as a bug in the software, which version is wrong (=which one is correct, and which result is caused by a buggy implementation)?

And remember: Integral-Calculator.com also provided two solutions for one input in a way similar to the example pictured above.

I posted this answer because it (at least in part) answers my question. This technically makes this answer wrong, but it's the least wrong answer available right now, so I want it to stay - at least until someone else provides a better one.

  • $\begingroup$ There are no unanswered questions. Your issue essentially amounts to the idea that $b$ and $b+1-1$, say, are different because they look different. $\endgroup$
    – Paul
    May 15, 2020 at 21:20
  • $\begingroup$ @Paul I believe that $ln(|\pi x -42 |)+C$ does not equal $ln(|x - \frac{42}{\pi}|) +C$. $\endgroup$
    – Tanja
    May 16, 2020 at 6:56
  • 3
    $\begingroup$ @Bernd-L: A way to avoid confusion about when $f(x)+C$ is equivalent to $g(x)+C$ is to see if $f(x)-g(x)$ is a constant. For $f(x)=\ln|\pi x-42|$ and $g(x)=\ln|x-42/\pi|$, we can write $$\begin{align}\ln|\pi x-42|-\ln|x-42/\pi|&=\ln\left(\pi\cdot|x-42/\pi|\right)-\ln|x-42/\pi|\\ &=\ln\pi+\ln|x-42/\pi|-\ln|x-42/\pi|\\ &=\ln\pi\quad\text{(const)}\end{align}$$ In the "$+C$" context, we have $g(x)+C=\ln\left|x-42/\pi\right|\color{red}{+C}$ and $f(x)+C=\ln|x-42/\pi|\color{red}{+\ln\pi+C}$; we allow the latter "$+C$" to absorb the $\ln\pi$, whereupon the two expressions are seen to be equivalent. $\endgroup$
    – Blue
    May 16, 2020 at 7:28

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