$\int\limits^\frac{\pi}{3}_0 {\frac{1}{cos^2x}+ \frac{xsinx}{cos^2x}} \, dx$
=$tanx|^\frac{\pi}{3}_0+\int\limits^\frac{\pi}{3}_0 { x} \, d(\frac{1}{cosx})$
=$\sqrt{3}+\frac{x}{cosx}|^\frac{\pi}{3}_0-\int\limits^\frac{\pi}{3}_0 {\frac{1}{cosx} } \, d(x)$
=$\sqrt{3}+\frac{2\pi}{3}-\int\limits^\frac{\pi}{3}_0 {\frac{cosx}{cos^2x} } \, d(x)$
=$\sqrt{3}+\frac{2\pi}{3}-\int\limits^\frac{\pi}{3}_0 {\frac{1}{1-sin^2x} } \, d(sinx)$
=$\sqrt{3}+\frac{2\pi}{3}-\int\limits^\frac{\pi}{3}_0 {\frac{1}{(1-sinx)(1+sinx)} } \, d(sinx)$
=$\sqrt{3}+\frac{2\pi}{3}-\int\limits^\frac{\pi}{3}_0 {\frac{1}{2}.(\frac{1}{1-sinx}+\frac{1}{1+sinx}) } \,d(sinx)$
=$\sqrt{3}+\frac{2\pi}{3}-\frac{1}{2}(-ln(1-sinx)+ln(1+sinx))|^\frac{\pi}{3}_0$
=$\sqrt{3}+\frac{2\pi}{3}+\frac{1}{2}(ln(1-sinx)-ln(1+sinx))|^\frac{\pi}{3}_0$
=$\sqrt{3}+\frac{2\pi}{3}+\frac{1}{2}ln\frac{1-sinx}{1+sinx}|^\frac{\pi}{3}_0$
=$\sqrt{3}+\frac{2\pi}{3}+\frac{1}{2}ln(7-4\sqrt{3})$
