The value of definite integral `int_0^(pi^2/4) dx/(1+sin sqrtx+ cos sqrtx)=` is

Answer

To solve the definite integral

\[ I = \int_0^{\frac{\pi^2}{4}} \frac{dx}{1 + \sin(\sqrt{x}) + \cos(\sqrt{x})} \]

we will perform a series of substitutions and manipulations. Let's go through the solution step by step.

### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, \( x = t^2 \) and \( dx = 2t \, dt \). The limits change as follows:
- When \( x = 0 \), \( t = 0 \).
- When \( x = \frac{\pi^2}{4} \), \( t = \frac{\pi}{2} \).

Thus, the integral becomes:

\[ I = \int_0^{\frac{\pi}{2}} \frac{2t \, dt}{1 + \sin(t) + \cos(t)} \]

### Step 2: Symmetry in the Integral
Now, we will consider the integral \( I \) and another integral \( J \) defined as:

\[ J = \int_0^{\frac{\pi}{2}} \frac{2\left(\frac{\pi}{2} - t\right) dt}{1 + \sin(t) + \cos(t)} \]

This represents the same integral but with \( t \) replaced by \( \frac{\pi}{2} - t \).

### Step 3: Simplifying \( J \)
Calculating \( J \):

\[ J = \int_0^{\frac{\pi}{2}} \frac{2\left(\frac{\pi}{2} - t\right) dt}{1 + \sin(t) + \cos(t)} = \int_0^{\frac{\pi}{2}} \frac{\pi - 2t}{1 + \sin(t) + \cos(t)} dt \]

### Step 4: Adding \( I \) and \( J \)
Now we add \( I \) and \( J \):

\[ I + J = \int_0^{\frac{\pi}{2}} \frac{2t \, dt}{1 + \sin(t) + \cos(t)} + \int_0^{\frac{\pi}{2}} \frac{\pi - 2t \, dt}{1 + \sin(t) + \cos(t)} \]

Combining these integrals gives:

\[ I + J = \int_0^{\frac{\pi}{2}} \frac{\pi \, dt}{1 + \sin(t) + \cos(t)} \]

### Step 5: Finding \( I \)
Since \( I = J \), we have:

\[ 2I = \int_0^{\frac{\pi}{2}} \frac{\pi \, dt}{1 + \sin(t) + \cos(t)} \]

Thus,

\[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\pi \, dt}{1 + \sin(t) + \cos(t)} \]

### Step 6: Evaluating the Integral
Now we need to evaluate the integral:

\[ \int_0^{\frac{\pi}{2}} \frac{dt}{1 + \sin(t) + \cos(t)} \]

Using the identity \( 1 + \sin(t) + \cos(t) = 2 + 2\sin\left(\frac{t + \frac{\pi}{4}}{2}\right) \) and simplifying, we can find:

\[ \int_0^{\frac{\pi}{2}} \frac{dt}{1 + \sin(t) + \cos(t)} = \frac{\pi}{2} \log(2) \]

### Final Result
Thus, substituting back, we find:

\[ I = \frac{1}{2} \cdot \frac{\pi}{2} \log(2) = \frac{\pi}{4} \log(2) \]

### Conclusion
The value of the definite integral is:

\[ \boxed{\frac{\pi}{4} \log(2)} \]