another problem

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- Dec 7th 2006, 04:06 PMtotalnewbieAnother integral
another problem

- Dec 7th 2006, 05:11 PMalinailiescu
- Dec 7th 2006, 05:20 PMtopsquark
Didn't you post this earlier? Hmmm...Might have been someone else.

Anyway...

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

Let $\displaystyle x = \frac{1}{t}$. Then $\displaystyle dx = -\frac{dt}{t^2}$.

$\displaystyle \int \frac{dx}{x \sqrt{x^2 - 1}} = \int \frac{ -\frac{dt}{t^2}}{ \frac{1}{t} \sqrt{ \frac{1}{t^2} - 1 } }$ $\displaystyle = - \int \frac{dt}{t \sqrt{\frac{1}{t^2} - 1}} = - \int \frac{dt}{\sqrt{t^2 \left ( \frac{1}{t^2} - 1 \right ) }}$

= $\displaystyle \int \frac{dt}{\sqrt{1 - t^2}}$

Do you recognize this integral?

-Dan