The value of the integral $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} dx$$ is:

Evaluate $$\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x}\, dx$$.

Let $$u = x + \frac{\pi}{4}$$ so that $$du = dx$$, and when $$x = -\frac{\pi}{4}$$ one has $$u = 0$$ whereas $$x = \frac{\pi}{4}$$ corresponds to $$u = \frac{\pi}{2}$$. Also $$x = u - \frac{\pi}{4}$$ implies $$\cos 2x = \cos\bigl(2u - \frac{\pi}{2}\bigr) = \sin 2u$$, thus the integral becomes $$I = \int_0^{\pi/2} \frac{u}{2 - \sin 2u}\, du$$.

Applying the identity $$\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$$ with $$a = \frac{\pi}{2}$$ gives $$I = \int_0^{\pi/2} \frac{\frac{\pi}{2} - u}{2 - \sin(\pi - 2u)}\, du = \int_0^{\pi/2} \frac{\frac{\pi}{2} - u}{2 - \sin 2u}\, du\,. $$ Adding these two expressions yields $$2I = \int_0^{\pi/2} \frac{\frac{\pi}{2}}{2 - \sin 2u}\, du = \frac{\pi}{2}\int_0^{\pi/2} \frac{du}{2 - \sin 2u}\,. $$

Set $$J = \int_0^{\pi/2} \frac{du}{2 - \sin 2u}$$ and use the Weierstrass substitution $$t = \tan u$$ so that $$\sin 2u = \frac{2t}{1+t^2}$$ and $$du = \frac{dt}{1+t^2}$$. This transforms the integral to $$J = \int_0^{\infty} \frac{1}{2 - \frac{2t}{1+t^2}} \cdot \frac{dt}{1+t^2} = \int_0^{\infty} \frac{dt}{2(1+t^2) - 2t} = \int_0^{\infty} \frac{dt}{2t^2 - 2t + 2} = \frac{1}{2}\int_0^{\infty} \frac{dt}{t^2 - t + 1}\,. $$

Completing the square in the denominator gives $$t^2 - t + 1 = \bigl(t - \tfrac12\bigr)^2 + \tfrac34$$ so that $$J = \frac{1}{2}\int_0^{\infty} \frac{dt}{\bigl(t - \tfrac12\bigr)^2 + (\tfrac{\sqrt{3}}{2})^2} = \frac{1}{2}\cdot\frac{1}{\frac{\sqrt{3}}{2}}\Bigl[\arctan\frac{t - \frac12}{\frac{\sqrt{3}}{2}}\Bigr]_0^{\infty} = \frac{1}{\sqrt{3}}\Bigl[\frac{\pi}{2} - \arctan\bigl(-\frac{1}{\sqrt{3}}\bigr)\Bigr] = \frac{2\pi}{3\sqrt{3}}\,. $$

Therefore $$2I = \frac{\pi}{2}\cdot\frac{2\pi}{3\sqrt{3}} = \frac{\pi^2}{3\sqrt{3}}\,, $$ and hence $$I = \frac{\pi^2}{6\sqrt{3}} = \frac{\pi^2\sqrt{3}}{18}\,. $$