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?
 A: The function given by Wolfram Alpha is somewhat problematic in that is contains both $\sqrt{x^3-4}$ and $\sqrt{4-x^3}$, so it unnecessarily brings in complex trig identities.
Hint: One great thing about $\frac{\mathrm{d}x}x$ is that the substitution $x\mapsto x^a$ maps $\frac{\mathrm{d}x}x\mapsto a\,\frac{\mathrm{d}x}x$. Thus, setting $x=u^{2/3}$ gives a standard trig substitution on $u$.
Full answer: if you want to peek, mouse over

$$ \begin{align} \int\frac{\sqrt{x^3-4}}{x}\,\mathrm{d}x &=\frac23\int\frac{\sqrt{u^2-4}}{u}\,\mathrm{d}u\\ &=\frac43\int\frac{\tan(\theta)}{\sec(\theta)}\tan(\theta)\sec(\theta)\,\mathrm{d}\theta\\ &=\frac43\int\tan^2(\theta)\,\mathrm{d}\theta\\ &=\frac43\int\left(\sec^2(\theta)-1\right)\mathrm{d}\theta\\ &=\frac43(\tan(\theta)-\theta)+C\\ &=\frac43\left(\sqrt{\frac{u^2}4-1}-\sec^{-1}\left(\frac u2\right)\right)+C\\&=\frac43\left(\sqrt{\frac{x^3}4-1}-\sec^{-1}\left(\frac{x^{3/2}}2\right)\right)+C \end{align} $$

A: $HINT$ : let $u-4 = t^2$, so $du = 2t\,dt$ and $u = t^2+4$. So it's easy to see where $\arctan$ comes.
A: \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.
