It seems that you are asking about double integrals, though the equations and expressions you've provided are a bit unclear. Let me break them down and attempt to clarify the tasks:

1. First Integral:

You seem to be describing a double integral in the following form: $\int \int_D \frac{x}{1 + x^2 + y^2} \, dx \, dy,$ where $D$ is the region defined by $0 \leq x \leq 1$, $0 \leq y \leq 1$, and $x^2 + y^2 \geq 1$.

• The region $D$ is bounded by the square $0 \leq x \leq 1$ and $0 \leq y \leq 1$, but it is also constrained by the condition $x^2 + y^2 \geq 1$.
• This suggests that we are integrating within the upper right quadrant of a circle of radius 1, intersecting the square.

This double integral is a bit tricky due to the constraint $x^2 + y^2 \geq 1$, and you would need to carefully determine the bounds for $y$ given each $x$. I'll need to figure out the exact setup for this region.

2. Second Integral:

The second expression seems to describe a double integral in the form: $\int \int_D x \cos(y) \, dx \, dy,$ where the region $D$ is bounded by:

• $y = 0$ (the x-axis),
• $y = x^2$ (a parabola opening upwards),
• $x = 2$ (a vertical line).

This integral appears more straightforward once we understand the region and set the bounds for $x$ and $y$.

Solution Approach:

1. For the first integral, the region $D$ requires careful consideration due to the circular constraint $x^2 + y^2 \geq 1$. You would:

   • Analyze the integral within the square, excluding the portion that lies inside the unit circle.
   • Set up the integral based on the limits for $x$ and $y$.

2. For the second integral, the bounds are simpler:

   • You integrate $x$ from 0 to 2.
   • For each $x$, $y$ varies from 0 to $x^2$.

Would you like me to work through the detailed solution for these integrals, including setting up the bounds and performing the integrations? Feel free to confirm!