Step 1: {Apply symmetry property of definite integrals}
Let:
\[
I = \int_{0}^{\pi/4} \frac{x}{1+\cos 2x + \sin 2x} \, dx.
\]
Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \), we get:
\[
I = \int_{0}^{\pi/4} \frac{\pi/4 - x}{1+\cos 2x + \sin 2x} \, dx.
\]
Step 2: {Combine integrals}
Adding the two forms of \( I \):
\[
2I = \int_{0}^{\pi/4} \frac{\pi/4}{1+\cos 2x + \sin 2x} \, dx.
\]
Step 3: {Simplify the integrand}
Rewrite \( 1 + \cos 2x + \sin 2x \):
We rewrite the denominator.
\[
1 + \cos 2x + \sin 2x = 1 + (2\cos^2 x - 1) + 2\sin x \cos x = 2\cos^2 x + 2\sin x \cos x = 2\cos x (\cos x + \sin x).
\]
The integral becomes:
\[
2I = \int_{0}^{\pi/4} \frac{\pi/4}{2\cos x (\cos x + \sin x)} \, dx = \frac{\pi}{8} \int_{0}^{\pi/4} \frac{1}{\cos^2 x + \sin x \cos x} \, dx.
\]
Dividing the numerator and denominator by \( \cos^2 x \):
\[
\frac{\pi}{8} \int_{0}^{\pi/4} \frac{\sec^2 x}{1 + \tan x} \, dx.
\]
Step 4: {Integrate and simplify}
Let \( u = 1 + \tan x \), so \( du = \sec^2 x \, dx \).
When \( x = 0 \), \( u = 1 + \tan 0 = 1 \).
When \( x = \pi/4 \), \( u = 1 + \tan(\pi/4) = 1 + 1 = 2 \).
The integral becomes:
\[
\frac{\pi}{8} \int_{1}^{2} \frac{1}{u} \, du = \frac{\pi}{8} [\log |u|]_{1}^{2} = \frac{\pi}{8} (\log 2 - \log 1) = \frac{\pi}{8} \log 2.
\]
Therefore, \( 2I = \frac{\pi}{8} \log 2 \), which means \( I = \frac{\pi}{16} \log 2 \).
Conclusion: The integral evaluates to \( \frac{\pi}{16} \log 2 \).