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.
Best Answer
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
$\sqrt{a^2-b}=u-v$
and adding their squares yields
$a=u+v.$
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.