# Integration and natural log question

WolframAlpha is confusing me. I'm working on two different integral problems, and with each one, for some reason, Wolfram switches the signs of the answer.

1. http://www.wolframalpha.com/input/?i=integral+dx/(x^.5(x-1))
2. http://www.wolframalpha.com/input/?i=integral+(x-1)/(x^2-4)

For the first one, they get $\log[1 - \sqrt{x}] - \log[(1 + \sqrt{x})]$ (see under alternate forms, thats the one I'm using).

However, I get: $\log[\sqrt{x}-1] - \log[(\sqrt{x}+1)]$ And so it differs by a sign, but its really bothering me, since I'm sure I'm doing it right. And the same applies to the second one:

They get: $3/4 \log(-x-2)+1/4 \log(x-2)$

I get: $3/4 \log(x+2)+1/4 \log(x-2)$

And so it differs by a sign again. (If this is a common mistake, then maybe you can point it out, but if this is just me and you can't see where I could have gone wrong, I can list my steps)

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1. The (real) indefinite integral is actually different in $(0, 1)$ and in $(1, \infty)$. WolframAlpha gave you the answer in the first interval and your answer works for the second. (Remember that in the real numbers, the logarithm of a non-positive number is undefined.)

2. This time WolframAlpha's answer is not defined for any real $x$. Your answer works in $(2, \infty)$ but there is a different antiderivative in $(-2, 2)$ and yet another antiderivative in $(-\infty, -2)$.

This is a fairly subtle property of the logarithm. Remember that any two antiderivatives of a function differ by a constant. The problem in this case is that the constant is complex, so if you aren't working in the complex numbers you get genuinely different answers.

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So my answer is correct then? I just wasted an hour trying to figure this out :( –  maq Dec 9 '10 at 23:16
That depends on which interval the question is asking you to find the antiderivative on. (Possibly your teacher or textbook is not making this distinction. That is unfortunate.) –  Qiaochu Yuan Dec 9 '10 at 23:31
I dont want to post another question for this, since it might be very similar, but can you explain why wolframalpha.com/input/?i=lim+n+approaches+infinity+of+ln((n^2%2Bn-1)‌​/(3n^2)) is equal to -log3? I get ln(1/3). (not sure how to fix the link in here, but make sure you copy that entire line because it cuts off) –  maq Dec 9 '10 at 23:34
-log 3 is equal to log (1/3). –  Qiaochu Yuan Dec 9 '10 at 23:39
@f-Prime: the ^ breaks links and doesn't get escaped to %5E by the site. If you change it manually, the link will work. (but it's too late for your comment) –  Ross Millikan Dec 10 '10 at 0:15