[b]Evaluate[/b]

 $\int^{\pi/2}_0 \frac{\cos^4x + \sin x \cos^3 x + \sin^2x\cos^2x + \sin^3x\cos x}{\sin^4x + \cos^4x + 2\ sinx\cos^3x + 2\sin^2x\cos^2x + 2\sin^3x\cos x} dx$
1. **Simplify the integrand:**

   The given integral is:
   \[
   I = \int_0^{\frac{\pi}{2}} \frac{\cos^4 x + \sin x \cos^3 x + \sin^2 x \cos^2 x + \sin^3 x \cos x}{\sin^4 x + \cos^4 x + 2 \sin x \cos^3 x + 2 \sin^2 x \cos^2 x + 2 \sin^3 x \cos x} \, dx
   \]

   Let's simplify the numerator and the denominator separately.

   **Numerator:**
   \[
   \cos^4 x + \sin x \cos^3 x + \sin^2 x \cos^2 x + \sin^3 x \cos x
   \]
   Factor out common terms:
   \[
   = \cos^2 x (\cos^2 x + \sin x \cos x) + \sin^2 x (\cos^2 x + \sin x \cos x)
   \]
   \[
   = (\cos^2 x + \sin^2 x)(\cos^2 x + \sin x \cos x)
   \]
   Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\):
   \[
   = (\cos^2 x + \sin x \cos x)
   \]

   **Denominator:**
   \[
   \sin^4 x + \cos^4 x + 2 \sin x \cos^3 x + 2 \sin^2 x \cos^2 x + 2 \sin^3 x \cos x
   \]
   Group terms:
   \[
   = (\sin^4 x + \cos^4 x + 2 \sin^2 x \cos^2 x) + (2 \sin x \cos^3 x + 2 \sin^3 x \cos x)
   \]
   Notice that:
   \[
   \sin^4 x + \cos^4 x + 2 \sin^2 x \cos^2 x = (\sin^2 x + \cos^2 x)^2 = 1
   \]
   And:
   \[
   2 \sin x \cos^3 x + 2 \sin^3 x \cos x = 2 \sin x \cos x (\cos^2 x + \sin^2 x) = 2 \sin x \cos x
   \]
   So the denominator simplifies to:
   \[
   = 1 + 2 \sin x \cos x
   \]

2. **Rewrite the integral:**

   Now the integral becomes:
   \[
   I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x + \sin x \cos x}{1 + 2 \sin x \cos x} \, dx
   \]

3. **Symmetry and substitution:**

   Notice that:
   \[
   \cos^2 x + \sin x \cos x = \sin^2 x + \sin x \cos x
   \]
   Therefore, we can write:
   \[
   I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x + \sin x \cos x}{1 + 2 \sin x \cos x} \, dx
   \]

4. **Combine the integrals:**

   Adding the two forms of the integral:
   \[
   2I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x + \sin x \cos x + \sin^2 x + \sin x \cos x}{1 + 2 \sin x \cos x} \, dx
   \]
   Simplify the numerator:
   \[
   = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x + \sin^2 x + 2 \sin x \cos x}{1 + 2 \sin x \cos x} \, dx
   \]
   Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\):
   \[
   = \int_0^{\frac{\pi}{2}} \frac{1 + 2 \sin x \cos x}{1 + 2 \sin x \cos x} \, dx
   \]
   \[
   = \int_0^{\frac{\pi}{2}} 1 \, dx
   \]
   \[
   = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2}
   \]

5. **Solve for \(I\):**

   \[
   2I = \frac{\pi}{2}
   \]
   \[
   I = \frac{\pi}{4}
   \]

The final answer is \(\boxed{\frac{\pi}{4}}\).