I'm trying to evaluate
$$\int \ \frac{2}{{}x\sqrt{9x^2 - 25}} dx$$
So I know that if I had just $$\int \ \frac{2}{{}x\sqrt{9x^2 - 25}} dx$$ then I would be able to use a natural log rule $\frac1a \text{arcsec}\left|\frac xa\right|+ c$ Am I able to pull out the $2$ from the integral? It would look like
$$2 \int \ \frac{1}{{}x\sqrt{9x^2 - 25}} dx$$
Furthermore, I have two $x$ values here $1$ and $9$, so would I have to further simplify in order to just have one $x$ value?
I could ALSO just do $\dfrac{2}{(x)(3x-5)}$ but does this make finding the integral harder or easier?
Just new at this guys and would love some clarification.
One standard way is to let $3x=5\sec \theta$. Then $3\,dx=5\tan\theta\sec\theta$. And then a miracle occurs. Our integral becomes $$\int \frac{6}{5\sec\theta\tan\theta}\cdot \frac{5}{3}\tan\theta\sec\theta\,d\theta,$$ Almost everything cancels. The integral is $2\theta+C$, and $\theta=\text{arcsec}(3x/5)$.
Another way is to rewrite our integrand as $\dfrac{2x}{x^2\sqrt{9x^2-25}}$. Now let $9x^2-25=u^2$. Then $9x\,dx=u\,du$ and $x^2=\dfrac{25+u^2}{9}$. So we end up with $$\int \frac{9}{(25+u^2)u}\cdot \frac{2u}{9}\,du.$$ Note the cancellation. The integral that remains is straightforward.