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

Can someone help?

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- Feb 7th 2010, 06:03 AMJosh146How do I find this integral?
$\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2-2x}} \ dx$

Can someone help? - Feb 7th 2010, 06:11 AMshawsend
How about complete the square in the radical then do one of those trig substitutions on (x-a)^2+b right?

- Feb 7th 2010, 06:16 AMJosh146