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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 Answer 1

up vote 3 down vote accepted
  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

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