# Integral from MIT Integration Bee 2023 Finals - $\int_{-1/2}^{1/2} \sqrt{x^2+1+\sqrt{x^4+x^2+1}}\,\textrm{d}x$

This question is from the MIT Integration Bee 2023 Finals, and this is Question 3. This integral was to be solved within four minutes, and the goal is to show that $$\int_{-1/2}^{1/2} \sqrt{x^2+1+\sqrt{x^4+x^2+1}}\ \textrm{d}x = \frac{\sqrt{7}}{2\sqrt{2}} + \frac{3}{4\sqrt{2}} \log\left(\frac{\sqrt{7}+2}{\sqrt{3}}\right)$$

My first attempt was to rewrite the inside of the nested square root as $$(x^2+1)^2 - x^2$$, after which I performed the trigonometric substitution $$x = \tan(\theta)$$. That made the integral above transform to $$\int_{-\alpha}^{\alpha} \sqrt{\sec^2(\theta)+\sqrt{\sec^4(\theta)-\tan^2(\theta)}}\sec^2(\theta)\ \textrm{d}\theta$$

Here, $$\alpha = \tan^{-1}(\frac{1}{2})$$. From here, I attempted to force the inside of the nested square root into some $$(a\pm b)^2$$ form, but doing so required me to be in $$\operatorname{GF}(2)$$. In my assumption, that meant $$\sec^4(\theta) - \tan^2(\theta) =(\sec^2(\theta) + \sec(\theta) + 1)^2$$, and the integral would transform to $$\int_{-\alpha}^{\alpha} \sqrt{2\sec^2(\theta) + \sec(\theta) + 1}\sec^2(\theta)\ \textrm{d}\theta$$

I'm pretty sure I made a mistake somewhere, but I don't know where. In the event I haven't, how could I simplify the inside of the nested square root? Or, what are other methods on attacking this question? I don't know complex analysis.

• I don't expect this could be solved using complex analysis anyway. I think you have to make a substitution of $x^2+1$. I will come back later with some more updates. Great question though! Feb 27 at 23:30
• Update: Integral Calculator couldn't find the indefinite integral. Wolfram Alpha couldn't find the definite integral. Feb 27 at 23:32
• Shocking! The nested square root extraction, which reduces the integral to familiar forms, should be something those calculators could check. Feb 28 at 0:13
• @OscarLanzi. Mathematica provides instantly the antiderivative and the result.. Even knowing it, I did not see what is the proper substitution. And they have 4 minutes to do it ? Feb 28 at 11:00
• @ClaudeLeibovici They did indeed have only four minutes. When looking at the video of it being done here, neither competitor answered correctly. It's of note that comprehensive work was not required to be shown. Feb 28 at 17:24

They pulled off a sneaky move.

Given a radical having the form

$$\sqrt{a+\sqrt{b}},$$

we may render

$$\sqrt{a+\sqrt{b}}=\sqrt{u}+\sqrt{v}$$

$$\sqrt{a-\sqrt{b}}=\sqrt{u}-\sqrt{v}$$

Multiplying these together gives

$$\sqrt{a^2-b}=u-v$$

$$a=u+v.$$

So if $$a^2-b=r^2$$ for some rational quantity $$r$$, we may render

$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+r}2}+\sqrt{\frac{a-r}2}$$

For the case at hand we find $$a=x^2+1,r=x$$ and so the integrand becomes

$$\sqrt{\frac{x^2+x+1}2}+\sqrt{\frac{x^2-x+1}2}.$$

The second term is just the first with $$x$$ exchanged for $$-x$$, so the antiderivative will be just the first term antiderivative minus (why?) its reflection (plus the usual arbitrary constant). The first term would then be integrated by usual methods of completing the square and trigonometric substitution.

• I understand up until the line "So if $a^2-b = r^2$ for some rational $r$" - why exactly can we say that? Feb 28 at 0:29
• I said "if". Meaning we have to check that. Denesting square roots dies not work otherwise. We find the condition holds in this case. Feb 28 at 0:32
• Great answer, for sure ! Feb 28 at 11:01
• Great solution. And could you please tell us the name of the contestant at MIT who pulled off this sneaky move? Mar 5 at 15:39
• I did not catch the names in the video. And the contestant did not make it all the way through in the alloted time. Mar 5 at 15:48

\begin{aligned} \because \quad x^2+1+\sqrt{x^4+x^2+1}&=\frac{1}{2}\left(2 x^2+2+2 \sqrt{\left(x^2+x+1\right) \left(x^2-x+1\right)}\right) \\ & =\frac{1}{2}\left(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\right) ^2 \\ \end{aligned} \begin{aligned}\therefore I&= \int_{-\frac 12}^{\frac 12} \sqrt{x^2+1+\sqrt{x^4+x^2+1}} \mathrm{~d} x\\&= 2\int_{0}^{\frac 12} \sqrt{x^2+1+\sqrt{x^4+x^2+1}} \mathrm{~d} x\\&= {\sqrt{2}} \int_0^{\frac{1}{2}}\left(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\right) d x \\&= { \sqrt{2}} \int_ {-\frac{1}{2}} ^{\frac{1}{2}}\sqrt{x^2+x+1}\ d x\\&= \boxed{\frac{1}{4 \sqrt{2}}\left[2 \sqrt{7}+3 \sinh ^{-1}\left(\frac{2}{\sqrt{3}}\right)\right] }\quad \textrm{(Refer to Footnote for details) } \end{aligned}

Footnote: \begin{aligned} J & =\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{x^2+x+1}\ d x=\frac{1}{2} \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{(2 x+1)^2+3}\ d x \end{aligned} Letting $$2x+1=\sqrt 3\sinh \theta$$ transforms $$J$$ into \begin{aligned} J & =\frac{3}{4} \int_{0}^{\sinh ^{-1} \frac{2}{\sqrt{3}} }\cosh ^2 \theta d \theta\\ & =\frac{3}{4} \int_0^{\sinh^{-1}\frac{2}{\sqrt{3}}} \frac{\cosh 2 \theta+1}{2} d \theta\\&= \frac{3}{8}\left[\frac{\sinh 2 \theta}{2}+\theta\right]_0^{\sinh ^{-1} \frac{2}{\sqrt{3}}} \\ & =\frac{3}{8}[\sinh \theta \cosh \theta+\theta]_0^{\sinh ^{-1} \frac{2}{\sqrt{3}}} \\ & =\frac{3}{8}\left(\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{7}}{\sqrt{3}}+\sinh ^{-1} \frac{2}{\sqrt{3}}\right) \\ &= \frac{1}{8}\left(2 \sqrt{7}+3 \sinh^{-1} \frac{2}{\sqrt{3}}\right) \end{aligned}