Solution of $\int\frac{\sqrt {x^3-4}}{x}dx$ : Need Hints $$\int\frac{\sqrt {x^3-4}} x \, dx$$
My attempt: $ \displaystyle \int\frac{3x^2\sqrt {x^3-4}}{3x^3}\,dx$
Then, substituting $u=x^3$; $\displaystyle \int\frac{\sqrt {u-4}}{3u} \, du$ 
$$\int\frac{u-4}{3u\sqrt{u-4}} \, du$$
$$\int\frac{1}{3\sqrt{u-4}}\,du-4\int\frac{1}{3u\sqrt{u-4}}\,du $$
I am having trouble with the 2nd part. And Wolfram Alpha says
Can you give me some hints on how to get arctanh function here?

\begin{align}
w & = \sqrt{u-4} \\
w^2 & = u-4 \\
2w\,dw & = du \\
w^2+4 & = u\\[15pt]
\int \frac{\sqrt {u-4}}{3u} \, du & = \int\frac{w}{3(w^2+4)} (2w\,dw) = \frac 2 3 \int \left( 1 - \frac 4 {w^2+4} \right) dw
\end{align}
etc.
