Let $I = \int\frac{2\cos 2x-1}{1+2\sin x}dx$

We know that $\cos 2x = 1 - 2\sin^2 x$. Substituting this into the integral, we get:

$I = \int\frac{2(1 - 2\sin^2 x)-1}{1+2\sin x}dx = \int\frac{2 - 4\sin^2 x - 1}{1+2\sin x}dx = \int\frac{1 - 4\sin^2 x}{1+2\sin x}dx$

We can rewrite the numerator as a difference of squares: $1 - 4\sin^2 x = (1 - 2\sin x)(1 + 2\sin x)$.

So, $I = \int\frac{(1 - 2\sin x)(1 + 2\sin x)}{1+2\sin x}dx = \int (1 - 2\sin x)dx$

$I = \int 1 dx - 2\int \sin x dx = x - 2(-\cos x) + C = x + 2\cos x + C$