The value of ∫-π/4^π/4(x+π/4)/(2-cos 2x) dx is

Split the numerator. The part $x/(2-\cos2x)$ is an odd function integrated over a symmetric interval, so it contributes zero. What remains is $\frac{\pi}{4}\int_{-\pi/4}^{\pi/4}\frac{dx}{2-\cos2x}=\frac{\pi}{2}\int_0^{\pi/4}\frac{dx}{2-\cos2x}$. Put $t=\tan x$, so $\cos2x=(1-t^2)/(1+t^2)$ and $dx=dt/(1+t^2)$: the integrand becomes $\dfrac{dt}{2(1+t^2)-(1-t^2)}=\dfrac{dt}{1+3t^2}$. Then $\int_0^{1}\frac{dt}{1+3t^2}=\frac{1}{\sqrt3}\tan^{-1}\sqrt3=\frac{\pi}{3\sqrt3}$. The integral equals $\frac{\pi}{2}\cdot\frac{\pi}{3\sqrt3}=\frac{\pi^2}{6\sqrt3}\approx0.9497$, which a numerical integration confirms. The option with $12\sqrt3$ forgets the doubling from symmetry.