Compute this integral (Is there a trick hidden to make it eassier?) I need some tips to compute this integral:
$$ \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx $$
What I did was express the denominator in the following form:
$$ \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx = \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{8x^2+x^2-1}}\,dx $$
Then, I made the change $x = \sec{\theta}$, then
$$ \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{8x^2+x^2-1}}\,dx = \int\,\dfrac{\sqrt{\sec^2{\theta}-1}}{\sec^5{\theta}\sqrt{8\sec^2{\theta}+\sec^2{\theta}-1}}\sec{\theta}\tan{\theta}\,d{\theta} $$
Trying to symplify this expression, I came to this:
$$ \dfrac{1}{4}\int\,\dfrac{\sin^2(2\theta)\cos{\theta}}{\sqrt{8\sin^2{\theta}+1}}\,d{\theta} $$
I feel this integral can be computed using some kind of "trick", but I can't see it. Thanks for your help and have a nice day!

You could try a partial fractional decomposition of the discriminant, and see if that approach is of any use. 
$$\frac{x^2-1}{9x^2-1} = \frac{1}{9}\cdot\left(1+\frac{4}{3x+1}-\frac{4}{3x-1}\right) \ .$$
