The problem and solution can be found here: Wolfram Alpha Solution (Click show steps tab to see work)

The method Wolfram Alpha uses to solve this integral involves many many sub steps. Are there any other easier ways to solve this integral or is this the only option?

Edit: The integral to be solved is $$\int \log{\sqrt{x^2-4}}\,dx.$$

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    $\begingroup$ You don't need to do anything fancy; just write it as $\frac{1}{2} \left( \log (x - 2) + \log (x + 2) \right)$. $\endgroup$ – Qiaochu Yuan Jul 27 '11 at 1:42
  • $\begingroup$ I deleted my answer since it was just a duplicate of that of Gerry. Btw even Wolfram is suggesting only integration by parts. Also don't think that the calculations are too long; they are quite straightforward. $\endgroup$ – Srivatsan Jul 27 '11 at 1:50
  • $\begingroup$ I deleted mine too (that makes three out of four answers deleted on this question :-) because it understated the simplification gained by Qiaochu's and Gerry's answer; my main point was, as Srivatsan wrote, that the W|A calculation isn't as involved as it looks; all the real work is done at the beginning and the rest is just trivial integrations and rearrangement. $\endgroup$ – joriki Jul 27 '11 at 3:06
  • $\begingroup$ Note that the W|A answer is rather inconvenient, as it includes a constant of $\pm\pi \mathrm i$ in the region $|x|\ge2$ where the integral is real. $\endgroup$ – joriki Jul 27 '11 at 3:20

$$\log\sqrt{x^2-4}=(1/2)\log(x^2-4),\qquad \log(x^2-4)=\log(x+2)+\log(x-2)$$

Now use integration by parts to do the two integrals that result.

  • $\begingroup$ Oh yea I forgot about log rules! doy! When 1/2 comes in front it come out of the integral right? $\endgroup$ – Matt Jul 27 '11 at 1:47
  • $\begingroup$ So it would end up being: (1/2)(int:log(x+2) + int:log(x-2)) right? $\endgroup$ – Matt Jul 27 '11 at 1:48
  • $\begingroup$ Yes, that's right. $\endgroup$ – Gerry Myerson Jul 27 '11 at 1:56

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