Proving $\int_0^1 \sqrt{x \left(\sqrt{-3 x^2+2 x+1}-x+1\right)} \, dx=\frac{7 \pi }{12 \sqrt{6}}$

Let $$y(x)= \sqrt{x \left(\sqrt{-3 x^2+2 x+1}-x+1\right)}\implies [y^2-x(1-x)]^2=x^2(1+2x-3x^2)$$

The integral $I=\int_0^1 y(x)dx$ represents the shaded area of the left graph. Rotate the curve with $x=\frac1{\sqrt2}(u+v)$ and $y=\frac1{\sqrt2}(u-v)$ for a symemtric functional form,

$$7u^4+12u^3v+26u^2v^2+12uv^3+7v^4=\sqrt2(6u^3+10u^2v+10uv^2+6v^3)$$

In polar coordinates $u=r\cos\theta$ and $v=r\sin\theta$, the function reads $$r(\theta)=\frac{2\sqrt2(\sin\theta+\cos\theta)(3+\sin2\theta)}{7+6\sin2\theta+3\sin^22\theta}$$

which is shown in the right graph. The equivalent area integral is,

$$I = \int_{-\pi/4}^{\pi/4} \frac12r^2(\theta)d\theta =\int_{0}^{\pi/2} \frac{8\cos^2t(3+\cos2t)^2}{(7+6\cos2t+3\cos^22t)^2}dt $$ where the variable change $t=\frac\pi4-\theta$ is made. Rewrite the integral with $\cos2t =\frac2{\sec^2t}-1$ and then let $u=\tan t$,

$$I = \int_{0}^{\pi/2} \frac{2(1+\sec^2t)^2\sec^2t}{(\sec^4t+3)^2}dt =\int_{0}^{\infty} \frac{2(u^2+2)^2}{(u^4+2u^2+4)^2}du$$ $$=\frac13\frac{u^3+u}{u^4+2u^2+4}\bigg|_0^\infty+\frac13\int_{0}^{\infty} \frac{u^2+5}{u^4+2u^2+4}du$$ $$=\frac7{12}\int_{0}^{\infty} \frac{2+u^2}{u^4+2u^2+4}du +\frac1{4}\int_{0}^{\infty} \frac{2-u^2}{u^4+2u^2+4}du =\frac7{12}\cdot \frac\pi{\sqrt6}+\frac1{4}\cdot 0=\frac{7\pi}{12\sqrt6}$$

Can be done in the same way as here, since $y - f(x)$ is (a part of) a rational curve. Taking $y = t x$ gives $$(x, y) = \left( \frac {2 (t^2 + 2)} {t^4 + 2 t^2 + 4}, \frac {2 t (t^2 + 2)} {t^4 + 2 t^2 + 4} \right), \\ \int_0^1 f(x) \, dx = -\int_0^\infty \dot x y \, dt.$$