# Calc II: Integration problem

1. Jan 12, 2010

### 2h2o

1. The problem statement, all variables and given/known data

Evaluate the indefinite integral:

2. Relevant equations

$$\int\frac{6}{x\sqrt{25x^2-1}} dx$$

3. 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!

2. Jan 12, 2010

### rock.freak667

Try a trig substitution x=cosecθ

3. Jan 12, 2010

### Altabeh

You're almost done: after substituting x^2=u^2+1/25 in the last integral you obtained, just bring 5 back into the integrand and then you need another substitution, basically, think of u=\sqrt(5)*cosh(x). Then you'll be left with a simple integral which is neccessarily required to be given a form like e^t/(1+e^2t) dt. From this point, you need one more substitution and then you are done by going back to x as you begin from this last one, pass t and u to definitely hit x again.

AB