Proving $\int_0^1 \sqrt{x \left(\sqrt{-3 x^2+2 x+1}-x+1\right)} \, dx=\frac{7 \pi }{12 \sqrt{6}}$ I was given that $$\int_0^1 \sqrt{x \left(\sqrt{-3 x^2+2 x+1}-x+1\right)} \, dx=\frac{7 \pi }{12 \sqrt{6}}.$$
Which is numerically correct. I tried a few substitutions, none of which lead to anything valuable. I'd like you to help me on the problem. Thanks in advance!


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