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$\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + 1 + \sin x}}{{1 + \cos 2x}}} dx = ?$

The given integral can be split as,

$ \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}} dx + \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{1 + \cos 2x}}} dx$

$\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}$ is an odd function. Thus, the given integral

$ = 0 + \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{2{{\cos }^2}x}}} dx = \frac{1}{2}\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {{{\sec }^2}xdx}  = \frac{1}{2}\left. {\tan x} \right|_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} = \frac{1}{{\sqrt 3 }}$