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!

$$I=\int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx$$
Let $u=\displaystyle\frac{\sqrt{x^2-1}}{\sqrt{9x^2-1}}$, which means that $\displaystyle\frac{du}{dx}=\frac{8x}{(x^2-1)^\frac{1}{2}(9x^2-1)^\frac{3}{2}}$
$$\Rightarrow I=\int\,\dfrac{u}{x^5}\,\frac{dx}{du}du$$
$$=\frac{1}{8}\int\,\dfrac{1}{x^6}\frac{\sqrt{x^2-1}}{\sqrt{9x^2-1}}\,\left((x^2-1)^\frac{1}{2}(9x^2-1)^\frac{3}{2}\right)du$$
$$=\frac{1}{8}\int\,\dfrac{(x^2-1)(9x^2-1)^2}{x^6}\,du$$
By some algebra you will find that $x=\displaystyle\frac{\sqrt{u^2-1}}{\sqrt{9u^2-1}}$, so
$$I=\frac{1}{8}\int\,\dfrac{\left(\frac{u^2-1}{9u^2-1}-1\right)\left(9\left(\frac{u^2-1}{9u^2-1}\right)-1\right)^2}{\left(\frac{u^2-1}{9u^2-1}\right)^3}\,du$$
which simplifies to
$$I=-64\int\,\dfrac{u^2}{(u^2-1)^3}\,du$$
Which can be found by the substitution $v=u^2-1$. This gives that 
$$I=2\ln\left(\frac{1+u}{1-u}\right)-\frac{8(u^3+u)}{(u^2-1)^2}$$ You should get an answer by back-substitution. Perhaps a trigonometric substitution like you suggested would have made it simpler but this seemed to work fine.
