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I have this integral:

$\int\dfrac{12x^3-24x^2+5}{x^2-2x}dx$

I solved using partial fractions and got an answer:

$6x^2-\frac{5}{2}\ln(x)+\frac{5}{2}\ln(x-2) +C$

But I am using "My Math Lab" online homework which is not accepting my answer as written above. I resolved it and got the same answer, and after some frustration, I plugged it into Wolfram Alpha and got this answer:

$6x^2-\frac{5}{2}\ln(x)+\frac{5}{2}\ln(2-x)+C$

So the only difference is $\ln(2-x)$ instead of what I wrote as $\ln(x-2)$.

What's more frustrating is MyMathLab still marked this wrong, and said the answer is actually:

$6x^2+\frac{5}{2}\ln\bigg|\frac{x-2}{x}\bigg|+C$

My first question is, why did Wolfram Alpha's answer say $\ln(2-x)$ instead of my original calculation that I found to be $\ln(x-2)$? Why is this switch occurring?

My second question is, why did MyMathLab mark mine and Wolfram Alpha's answer incorrect? Is this one of the limitations of MyMathLab grading, where two answers are actually the same and correct, but only look different? Or is Wolfram Alpha wrong? Why was the correct answer written the way it was? Where did $\frac{5}{2}\ln\bigg|\frac{x-2}{x}\bigg|$ come from?

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  • $\begingroup$ As indicated in the answer, it's preferable to use $\ln|x|$ and $\ln|x-2|$ in the answer, to get an antiderivative with the largest possible domain; and then $\frac{5}{2}(\ln|x-2|-\ln |x|)=\frac{5}{2}\ln|\frac{x-2}{x}|$. $\endgroup$ – user84413 Sep 3 '14 at 23:46
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The solutions have different domains.

Your answer is a valid solution for $x > 2$.

Wolfram's answer is a valid solution for $0<x<2$.

MyMathLab's answer is a valid solution for $x\in(-\infty,0)\cup(0,2)\cup(2,\infty)$ -- but it's not the only possible antiderivative on this domain, because it doesn't take into account that the integration constant is allowed to change at $x=0$ and $x=2$ where the integrand isn't defined.

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  • $\begingroup$ What math rules allow you to simply just change the contents of $\ln(x-2)$ to be $\ln(2-x)$? $\endgroup$ – Sabien Sep 3 '14 at 23:02
  • $\begingroup$ @Sabien: Who says there are such rules? Each of the solutions independently satisfies the criterion (of having the desired derivative) on their respective domains. They don't arise from each other by some rule of rewriting. $\endgroup$ – Henning Makholm Sep 3 '14 at 23:05

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