MIT Integration Bee 2023 - Finals - Problem 3

The problem:

Solution. We assume $x > 0$.

$$\begin{aligned}I &= \int \sqrt{x^2+1+\sqrt{x^4+x^2+1}} \, dx \\ &= \int \frac{t^2-t+1}{(2t-1)^{3/2}\sqrt{t-2}} \, dt & \left(t = x^2+1+\sqrt{x^4+x^2+1}\right) \\[1em] &= 2 \int \frac{u^4+3u^2+3}{(2u^2+3)^{3/2}} \, du & \left(u^2 = t-2\right) \\[1em] &= \frac{\sqrt{2}}{2} \int \frac{3s^8+12s^6+34s^4+12s^2+3}{8s^3(s^2+1)^2} \, ds & \left(\text{USM Transform 3: } u = \sqrt{\frac{3}{2}}\frac{s^2-1}{2s}\right) \\[1em] &= \frac{\sqrt{2}}{2} \int \left( \frac{3s}{8} + \frac{3}{4s} + \frac{3}{8s^3} + \frac{2s}{(s^2+1)^2} \right) \, ds & (\text{PFD}) \\[1em] &= \frac{\sqrt{2}}{2} \left( \frac{3s^2}{16} + \frac{3}{4}\ln|s| - \frac{3}{16s^2} - \frac{1}{s^2+1} \right) + C \\[1em]
&= \frac{3\sqrt{2}}{32}\left(s^2 - \frac{1}{s^2}\right) + \frac{3\sqrt{2}}{8}\ln|s| - \frac{\sqrt{2}}{2(s^2+1)} + C
\end{aligned}$$

Where:

$$s = \sqrt{\frac{2}{3}}u + \sqrt{\frac{2}{3}u^2+1}, \quad u = \sqrt{t-2}, \quad t = x^2+1+\sqrt{x^4+x^2+1}.$$

Using the symmetry of the even function $f(x)$, we calculate $2 \int_{0}^{1/2} f(x) \, dx$:

$$\begin{aligned}\text{Limits for } s: \quad & x=0 \implies s=1 \\
& x=1/2 \implies s = \sqrt{\frac{2+\sqrt{7}}{\sqrt{3}}}\end{aligned}$$

$$\begin{aligned}\int_{-1/2}^{1/2} f(x) \, dx &= 2 \left[ F(s) \right]_{1}^{\sqrt{\frac{2+\sqrt{7}}{\sqrt{3}}}} \\[1em] &= 2 \left[ \left( \frac{\sqrt{14}}{8} - \frac{\sqrt{2}}{4} + \frac{3\sqrt{2}}{16}\ln\left(\frac{2+\sqrt{7}}{\sqrt{3}}\right) \right) - \left( -\frac{\sqrt{2}}{4} \right) \right] \\[1em] &= 2 \left[ \frac{\sqrt{14}}{8} + \frac{3\sqrt{2}}{16}\ln\left(\frac{2+\sqrt{7}}{\sqrt{3}}\right) \right] \\[1em] &= \frac{\sqrt{14}}{4} + \frac{3\sqrt{2}}{8}\ln\left(\frac{2+\sqrt{7}}{\sqrt{3}}\right) \end{aligned}$$