Integrating $\int \dfrac{2x \ln x}{\sqrt{x^2-9}}\mathrm dx$ by parts. We have to integrate $$\int\frac{2x \ln x}{\sqrt{(x^2-9)}} \mathrm dx$$
Is it right to use Integration by Parts?
I tried to substitute it with
$$u = \log x,\: \mathrm du = \frac 1x \mathrm dx;$$
$$v = x^2 - 9,\: \mathrm dv = 2x \mathrm dx.$$
But then I'm stuck with substituting it within the original equation because from $\mathrm du = \dfrac 1x \mathrm dx,$ and $\mathrm dv = 2x \mathrm dx,$ there will be two $\mathrm dx$'s to substitute and from $\mathrm du = \dfrac 1x \mathrm dx,$ the $x$ will go to the denominator and I don't know what to do any more.

You need to make the integral into only two parts, $u$ and $dv.$ So you should let $u= \ln x$ and $dv = \dfrac{2x}{\sqrt{x^2-9}} dx.$ 
Then using those choices, work out what $du$ and $v$ must be, and put it all in the formula $\displaystyle \int u dv = uv - \int v du.$
