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$$\int\frac{\sqrt {x^3-4}} x \, dx$$

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 saysenter image description here

Can you give me some hints on how to get arctanh function here?

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  • $\begingroup$ You can simply assume x^3-4 as u ,then simplify it and proceed with multiplication rule . $\endgroup$ – 0xVikas Dec 12 '17 at 6:30
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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

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

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

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\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.

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  • 2
    $\begingroup$ I believe it should be $u=w^2\color{#C00}{+}4$. $\endgroup$ – robjohn Dec 12 '17 at 4:54

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