Example 6 (to $5^{\circ}$ ) Find $\int \frac{\sin 2 x d x}{1+\cos 2 x+\sin ^{2} 2 x}$.
Solution. We have $R(\sin 2 x, \cos 2 x)=\frac{\sin 2 x}{1+\cos 2 x+\sin ^{2} 2 x} \quad$ and $R(-\sin 2 x, \cos 2 x)=-R(\sin 2 x, \cos 2 x)$. Case 1$)$ applies. Therefore, let $\cos 2 x=t$. This substitution is equivalent to bringing $\cos 2 x$ under the differential sign. Therefore, we can do without additional formulas.

$$
\begin{aligned}
& \int \frac{\sin 2 x d x}{1+\cos 2 x+\sin ^{2} 2 x}=\frac{1}{2} \int \frac{d t}{t^{2}-t-2}=\frac{1}{2} \int \frac{1}{3}\left(\frac{1}{t-2}-\frac{1}{t+1}\right) d t= \\
& =\frac{1}{6} \ln \left|\frac{t-2}{t+1}\right|+C=\frac{1}{6} \ln \frac{2-\cos 2 x}{1+\cos 2 x}+C
\end{aligned}
$$

Note. This integral can also be represented as $\int \frac{\sin 2 x d x}{2-\cos ^{2} 2 x+\cos 2 x}=-\frac{1}{2} \int \frac{d u}{2+u-u^{2}}$, where $u=\cos 2 x$.