Even or Odd

Firstly $I \; = \; \int_{-\frac14\pi}^{\frac14\pi} \frac{\sin2x \ln(1 + e^x)}{(2 + \cos2x)^2}\,dx \; = \; \int_{-\frac14\pi}^{\frac14\pi} \frac{\sin2x}{(2 + \cos2x)^2}\big[\tfrac12x + \ln(e^{\frac12x} + e^{-\frac12x})\big]\,dx \; = \; \frac12\int_{-\frac14\pi}^{\frac14\pi} \frac{x\sin2x}{(2 + \cos2x)^2}\,dx$ since $\frac{\sin2x \ln(e^{\frac12x}+e^{-\frac12x})}{(2 + \cos2x)^2}$ is an odd function. Now integrating by parts gives $I \; = \; \frac14\left[ \frac{x}{2 + \cos2x}\right]_{-\frac14\pi}^{\frac14\pi} - \frac14\int_{-\frac14\pi}^{\frac14\pi} \frac{dx}{2 + \cos2x} \; = \; \frac{\pi}{16} - \frac14\int_{-\frac14\pi}^{\frac14\pi} \frac{dx}{2 + \cos2x} \; = \; \frac{\pi}{16} - \frac18\int_{-\frac12\pi}^{\frac12\pi} \frac{dx}{2 + \cos x}$ Finally, the substitution $t = \tan\tfrac12x$ gives $I \; = \; \frac{\pi}{16} - \frac{1}{4}\int_{-1}^1 \frac{dt}{3 + t^2} \; = \; \frac{\pi}{16} - \frac{1}{ 4\sqrt{3}}\left[\tan^{-1}\tfrac{t}{\sqrt{3}}\right]_{-1}^1 \; = \; \frac{\pi}{16} - \frac{\pi}{12\sqrt{3}} \; = \; \boxed{\frac{\pi}{144}(9 - 4\sqrt{3}) \; = \; 0.045199593829843926}$