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Homework Statement
Evaluate the indefinite integral:
Homework Equations
\int\frac{6}{x\sqrt{25x^2-1}} dx
The Attempt at a Solution
I've tried a number of different substitutions, but this last one gets me the closest, I think.
\int\frac{6}{x\sqrt{25x^2-1}} dx = \frac{6}{5}\int\frac{1}{x\sqrt{x^2-\frac{1}{25}}} dx = \frac{6}{5}\int\frac{x}{x^2\sqrt{x^2-\frac{1}{25}}} dx
let u=\sqrt{x^2-\frac{1}{25}
then
du=\frac{x dx}{\sqrt{x^2-\frac{1}{25}}}}}
This is where I want to substitute in, but it doesn't go. \frac{x}{\sqrt{x^2-\frac{1}{25}}}}}}dx fits in, except for the x^2 in the denominator which doesn't work with my substitution. I don't see where to go from here, or what I should have done differently.
Thanks!