integrate 1/(sqrt(x^2-4))

**MechEng** · June 30th 2010, 10:29 AM

Good afternoon,

I managed to work my way through this problem rather quickly, but wound up with an answer that I feel is not correct (at least according to my calculator).

I have:

$ \int{\frac{1}{\sqrt{x^2-4}}dx $

I evaluate to:

$ \ln\mid \frac{x+\sqrt{x^2-4}}{2}\mid + C $

It should be:

$ \ln\mid{x+\sqrt{x^2-4}}\mid + C $

I am picking up the 2 in the denominator via $sec\theta = \frac{x}{2}$ and $tan\theta = \frac{\sqrt{x^2-4}}{2}$ .

Can anyone tell me what it is that I'm doing wrong here?

**skeeter** · June 30th 2010, 10:55 AM

Originally Posted by MechEng

Good afternoon,

I managed to work my way through this problem rather quickly, but wound up with an answer that I feel is not correct (at least according to my calculator).

I have:

$ \int{\frac{1}{\sqrt{x^2-4}}dx $

I evaluate to:

$ \ln\mid \frac{x+\sqrt{x^2-4}}{2}\mid + C $

It should be:

$ \ln\mid{x+\sqrt{x^2-4}}\mid + C $

I am picking up the 2 in the denominator via $sec\theta = \frac{x}{2}$ and $tan\theta = \frac{\sqrt{x^2-4}}{2}$ .

Can anyone tell me what it is that I'm doing wrong here?

nothing wrong ...

$\displaystyle \ln\left|\frac{x+\sqrt{x^2-4}}{2}\right| + C = \ln\left|x + \sqrt{x^2-4}\right| - \ln{2} + C$

... which only differs by a constant from the "it should be" solution.

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