I try to integrate
$$\int \frac{1}{x\sqrt{9x^2-1}}\,dx$$
let $u=x^2,\quad \quad du=2x\,dx,\:\quad \:dx=\frac{1}{2x}\,du$
$$
\begin{align}
& \int \frac{1}{x\sqrt{9u-1}}\frac{1}{2x}\,du \\[8pt]
= {} & \int \frac{1}{2x^2\sqrt{9u-1}} \, du \\[8pt]
= {} & \frac{1}{2}\int \frac{1}{u\sqrt{9u-1}}\, du
\end{align}
$$
now let $v=\sqrt{9u-1},\quad \quad dv=\frac{9}{2\sqrt{9u-1}}du,\quad \:dv=\frac{9}{2v}du,\:\quad \:du=\frac{2v}{9}dv$
$$\frac{\int \frac{1}{uv}\frac{2v}{9}dv}{2} = \frac{\int \frac{1}{u}dv}{9}$$
let $v=\sqrt{9u-1},\:u=\frac{v^2+1}{9}$
$$\frac{\int \frac{9}{v^2+1}dv}{9} = \int \frac{dv}{v^2+1}$$
since $\int \frac{1}{v^2+1}\,dv=\arctan(v)$
substitute back $v$ and $u$, then i get
$$\arctan \left(\sqrt{9x^2-1}\right)$$
this is my answer, but i'm not sure if my answer is correct or not. Please if you have a better calculation than mine, i would be really happy if you want to show me and correct my answer. Thank you so much.
Best Answer
Hint
I suppose that you could go faster to the solution if you start changing variable $$\sqrt{9 x^2-1}=u$$ that is to say $$x=\frac{\sqrt{u^2+1}}{3}$$ $$dx=\frac{u}{3 \sqrt{u^2+1}} du$$ and then $$\int \frac{dx}{x\sqrt{9x^2-1}}=\int \frac{du}{u^2+1}$$ The whole idea was just to get rid of the radical as fast as possible.