\( \int _{}^{}\frac{\cos (2x)}{1+\cos (x)+\sin (x)} d x \)

To address the integral \( \int \frac{\cos(2x)}{1+\cos(x)+\sin(x)} \, dx \), we can start by simplifying the denominator. Notice that \( \cos(2x) = 2\cos^2(x) - 1 \), which might help us in our manipulation. We can also manipulate the denominator \( 1 + \cos(x) + \sin(x) \) by recognizing it could be expressed in terms of sine and cosine if we expand it further. One effective approach is to use a substitution or to transform the integral into a more manageable form. Consider the substitution \( t = \tan(\frac{x}{2}) \), which gives us the relationships \( \sin(x) = \frac{2t}{1+t^2} \) and \( \cos(x) = \frac{1-t^2}{1+t^2} \), as well as \( dx = \frac{2}{1+t^2} dt \). From this, we substitute and then express all terms in t's terms, simplifying the integral into a rational function. While the process is quite intricate, with careful algebra, one can resolve the resulting setup into simpler terms or even find a direct anti-derivative depending on how simplifications proceed. Ultimately, confirming convergence and calculating limits may also be necessary, especially when approaching more complex integrals like these.