To find the double integral of $\frac{1}{1 + xy}$ over the region where $x$ and $y$ both vary from 0 to 1, we set up the integral as follows:

$\iint_{0 \le x \le 1, \, 0 \le y \le 1} \frac{1}{1 + xy} \, dx \, dy$

We can change the order of integration because the region of integration is a square, and the integrand is continuous on this region. Hence, the double integral can be written as:

$\iint_{0 \le x \le 1, \, 0 \le y \le 1} \frac{1}{1 + xy} \, dx \, dy = \int_{0}^{1} \left( \int_{0}^{1} \frac{1}{1 + xy} \, dx \right) dy$

To solve the inner integral:

$\int_{0}^{1} \frac{1}{1 + xy} \, dx$

We perform a substitution. Let $u = 1 + xy$. Then $du = y \, dx$ or $dx = \frac{du}{y}$. When $x = 0$, $u = 1$. When $x = 1$, $u = 1 + y$. Thus, the integral becomes:

$\int_{0}^{1} \frac{1}{1 + xy} \, dx = \int_{1}^{1 + y} \frac{1}{u} \cdot \frac{1}{y} \, du = \frac{1}{y} \int_{1}^{1 + y} \frac{1}{u} \, du$

Evaluating the integral:

$\frac{1}{y} \left[ \ln u \right]_{1}^{1 + y} = \frac{1}{y} \left( \ln(1 + y) - \ln(1) \right) = \frac{\ln(1 + y)}{y}$

Now we need to integrate this result with respect to $y$:

$\int_{0}^{1} \frac{\ln(1 + y)}{y} \, dy$

To evaluate this integral, we use the fact that:

$\int_{0}^{1} \frac{\ln(1 + y)}{y} \, dy = \int_{0}^{1} \int_{0}^{y} \frac{dx}{1 + xy} \, dy$

Using the identity:

$\int_{0}^{1} \frac{\ln(1 + y)}{y} \, dy = \frac{\pi^2}{12}$

Thus, the value of the original double integral is:

$\iint_{0 \le x \le 1, \, 0 \le y \le 1} \frac{1}{1 + xy} \, dx \, dy = \frac{\pi^2}{12}$

Therefore, the double integral evaluates to $\boxed{\frac{\pi^2}{12}}$.

Do you want more details or have any questions?

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Here are 8 related questions that expand on this information:

1. How would you evaluate the double integral if the limits were different?
2. Can you explain the change of order of integration in double integrals?
3. What is the general method to solve double integrals with a substitution?
4. How does Fubini's theorem apply to this problem?
5. What are some applications of double integrals in physics?
6. How would you interpret the result geometrically?
7. Can this integral be solved using numerical methods? If so, how?
8. How does the integral change if the integrand is modified to $\frac{1}{1 + x^2 y^2}$?

Tip: When solving double integrals, consider if a change of variables or the order of integration can simplify the computation.