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I'm given $$\int \frac{x^3}{\sqrt{x^4+x^2+1}}dx$$

My attempt, Let $u=x^2$, $du=2xdx$

$$=\frac{1}{2}\int \frac{u}{\sqrt{u^2+u+1}}du = \frac{1}{2}\int \frac{u}{\sqrt{(u+\frac{1}{2})^2+\frac{3}{4}}}du$$

Let $s=u+\frac{1}{2}$, $ds=du$

$$=\frac{1}{2}\int\frac{s-\frac{1}{2}}{\sqrt{s^2+\frac{3}{4}}} ds$$

Let $s=\frac{\sqrt{3}}{2}\tan p$, $ds=\frac{\sqrt{3}}{2}\sec^2 p\,dp$

So, $$\sqrt{s^2+\frac{3}{4}}=\sqrt{\frac{3\tan^2 p}{4}+\frac{3}{4}} =\frac{\sqrt{3}}{2}\sec p$$

and $$p=\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)$$

$$=\frac{\sqrt{3}}{4}\int \frac{2\left(\frac{\sqrt{3}}{2}\tan p-\frac{1}{2}\right)\sec p}{\sqrt{3}}dp$$

$$=\frac{1}{2}\int \left(\frac{\sqrt{3}}{2}\tan p-\frac{1}{2}\right)\sec p\,dp$$

$$=\frac{\sqrt{3}}{4}\int \tan p\sec p\,dp-\frac{1}{4}\int \sec p\,dp$$

Let $w=\sec p$, $dw=\tan p\sec p\,dp$

$$=\frac{\sqrt{3}}{4}\int 1 dw-\frac{1}{4}\int \sec p\,dp$$

$$=\frac{\sqrt{3}w}{4}-\frac{1}{4}\int \sec p\,dp$$

$$=\frac{\sqrt{3}w}{4}-\frac{1}{4}\ln (\tan p+\sec p)+c$$

$$=\frac{1}{4}\sqrt{3}\sec p-\frac{1}{4}\ln (\tan p+\sec p)+c$$

$$=\frac{1}{4}\sqrt{3}\sec \left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)-\frac{1}{4}\ln \left[\tan\left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)+\sec \left(\tan^{-1}\left(\frac{2s}{\sqrt{3}}\right)\right)\right]+c$$

$$=\frac{1}{4}\sqrt{4s^2+3}+\frac{1}{8}\left[\ln 3 -2\ln \left(\sqrt{4s^2+3}+2s\right)\right]+c$$

$$=\frac{1}{2}\sqrt{u^2+u+1}+\frac{1}{8}\left[\ln 3-2\ln \left(2\sqrt{u^2+u+1}+2u+1\right)\right]+c$$

$$=\frac{1}{2}\sqrt{x^4+x^2+1}+\frac{1}{8}\left[\ln 3-2\ln \left(2x^2+2\sqrt{x^4+x^2+1}+1\right)\right]+c$$

$$=\frac{1}{8}\left[4\sqrt{x^4+x^2+1}-2\ln \left(2x^2+2\sqrt{x^4+x^2+1}+1\right)+\ln 3\right]+c$$

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  • $\begingroup$ What about writing $$\frac{1}{2}\int \frac{u}{\sqrt{u^2+u+1}}du=\frac{1}{2}\int \frac{u+\frac{1}{2}-\frac{1}{2}}{\sqrt{u^2+u+1}}du$$ and notice that the numerator is looking like the derivative of what is in the radical. This could make things shorter (I hope) $\endgroup$ – Claude Leibovici Apr 20 '15 at 9:13
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Some ideas:

$$\int\frac{x^3}{\sqrt{x^4+x^2+1}}dx=\frac14\int\frac{\overbrace{4x^3+2x}^{=(x^4+x^2+1)'}}{\sqrt{x^4+x^2+1}}dx-\frac14\overbrace{\int\frac{2xdx}{\sqrt{x^4+x^2+1}}}^{u:=x^2\implies du=2xdx}=$$

$$\frac12\sqrt{x^4+x^2+1}-\frac14\int\frac{du}{\sqrt{u^2+u+1}}=\frac12\sqrt{x^4+x^2+1}-\frac14\int\frac{du}{\sqrt{\left(u+\frac12\right)^2+\frac34}}=$$

$$=\frac12\sqrt{x^4+x^2+1}-\frac1{2\sqrt3}\int\frac{du}{\sqrt{\left(\frac{\sqrt3}2u+\frac{\sqrt3}4\right)^2+1}}=$$

$$=\frac12\sqrt{x^4+x^2+1}-\frac13\int\frac{\frac{\sqrt3}2du}{\sqrt{\left(\frac{\sqrt3}2u+\frac{\sqrt3}4\right)^2+1}}=$$

Finally, change variables: $\;\sinh t=\dfrac{\sqrt3}2u+\dfrac{\sqrt3}4\;,\;\;\cosh t\,dt=\dfrac{\sqrt3}2du\;$

and get a very, very simple last integral above...and then go back to the original variable if you wish.

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We have $$ I=\int \frac{x^3}{\sqrt{ x^4+x^2+1}} dx= \frac{1}{4} \int \frac{4x^3 +2x}{\sqrt{ x^4+x^2+1}} dx+ \frac{1}{4} \int \frac{-2x}{\sqrt{ x^4+x^2+1}} dx$$ $$ = \frac{1}{4} \int \frac{ dv}{\sqrt v} - \frac{1}{2} \int \frac{x dx}{\sqrt{( x^2+\frac{1}{2})^2+ \frac{3}{4}}} \quad \text{ I take the change of variable } v=x^4+x^2+1$$ Now, let $u=x^2+\frac{1}{2} $ implies $du= 2xdx$ , then substitute in the second integral to get: $$I=\frac{1}{2} \sqrt v -\frac{1}{4} \int \frac{ du}{\sqrt{u^2+ \frac{3}{4}}} =\frac{1}{2} \sqrt v -\frac{1}{4} \sinh^{-1}(\frac{2u}{\sqrt{3}}) $$ again to our variable $x$: $$I=\frac{1}{2} \sqrt { x^4+x^2+1} -\frac{1}{4} \sinh^{-1}(\frac{2x^2 +1}{\sqrt{3}}) + C $$ with simple simplification you will get your form(using $\sinh^{-1}(z)= \ln (z+\sqrt{z^2+1})$).

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  • $\begingroup$ I fixed some very minor things to get $\sinh^{-1}(z)$ instead. I hope you don't mind. Cheers :-) $\endgroup$ – Claude Leibovici Apr 20 '15 at 9:35
  • $\begingroup$ @ClaudeLeibovici Thank you for the editing :) $\endgroup$ – Nizar Apr 20 '15 at 9:42
  • $\begingroup$ You are very welcome ! In fact, it seems that it is almost standard nowdays to write $\cos^1{-1}(x)$ instead of $arccos(x)$ which was the standard when I was young (loooong time ago !). $\endgroup$ – Claude Leibovici Apr 20 '15 at 10:01
  • $\begingroup$ @ClaudeLeibovici in my university even some times we use to write $argcosh$ ! I dont know if it makes sense or not, but usually we use it ! $\endgroup$ – Nizar Apr 20 '15 at 10:03
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Substituting $\frac{2x^2+1}{\sqrt{3}}=u\implies x=\frac{\sqrt{\sqrt{3}u-1}}{\sqrt{2}}$, the integral becomes:

$$\begin{align} \int\frac{x^3}{\sqrt{x^4+x^2+1}}\,\mathrm{d}x &=\int\frac{\left(\sqrt{3}u-1\right)^{3/2}}{\sqrt{6}\sqrt{u^2+1}}\cdot\frac{\sqrt{6}\,\mathrm{d}u}{4\sqrt{\sqrt{3}u-1}}\\ &=\frac14\int\frac{\sqrt{3}u-1}{\sqrt{u^2+1}}\,\mathrm{d}u\\ &=\frac{\sqrt{3}}{8}\int\frac{2u\,\mathrm{d}u}{\sqrt{u^2+1}}-\frac14\int\frac{\mathrm{d}u}{\sqrt{u^2+1}}\\ &=...\\ \end{align}$$

This seems to me to be the most direct route to the antiderivative.

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