Wolfram gave me one answer, examiners gave me another

[Saibot]

Homework Statement

Find the integral of 1/(2x-2)

Relevant Equations

u substitution

I removed the coefficient of 1/2 before integrating. So I had:
1/2 integral[(1/x-1)]
= 1/2 ln(x-1) + C

Using u substitution without removing the coefficient yields
1/2 ln(2x-2) + C

I get how it works both ways, and my answer must be wrong, but what exactly is causing this conflict? Am I not supposed to remove the factor of 1/2?

What is happening here?

 

[Orodruin]

$\ln(2x-2) = \ln(2(x-1)) = \ln(x-1) + \ln 2$
Therefore, $\ln(x-1)$ and $\ln(2x-2)$ differ only by a constant so both (divided by 2) are primitive functions of $1/(2x-2)$.

 

[Delta2]

What is happening is that as @Orodruin explained both are antiderivatives (=indefinite integrals) of the given function, however you use in both expressions the same letter $C$ for the constant and that is what is causing the confusion. We know that $C$ can be anything so it doesn't matter if you have just $\ln x+C$ in one expression and $\ln x+C+\ln2$ in the other, because if $C$ is anything, then so is $C+\ln2$ anything.

 

[Saibot]

Got it, thanks guys. So I'm not actually wrong, we'll both just get different expressions for C.

 

[PeroK]

Got it, thanks guys. So I'm not actually wrong, we'll both just get different expressions for C.

It's happens quite often with integration. For example:
$$\cos(2x) = \cos^2 x - \sin^2 x = 2\cos^2 x -1 = 1 - 2\sin^2 x$$Any one of these may result depending on how you tackle the integral $$-2 \int \sin(2x)\ dx$$