Suppose that $f(x, y)=\frac{y}{1+x}$ at which $\{(x, y) \mid 0 \leq x \leq 1,-x \leq y \leq \sqrt{x}\}$. Then the double integral of $f(x, y)$ over $D$ is \[ \iint_{D} f(x, y) d x d y= \] Round your answer to four decimal places.

Step 1 ：Define the function \(f(x, y)=\frac{y}{1+x}\) over the region \(D\) defined by the inequalities \(0 \leq x \leq 1\) and \(-x \leq y \leq \sqrt{x}\).

Step 2 ：First, integrate \(f(x, y)\) with respect to \(y\), with limits of integration from \(-x\) to \(\sqrt{x}\). This gives the intermediate result \(-\frac{x^2}{2x + 2} + \frac{x}{2x + 2}\).

Step 3 ：Next, integrate this result with respect to \(x\), with limits of integration from 0 to 1. This gives the final result \(\frac{3}{4} - \ln(2)\).

Step 4 ：Thus, the double integral of \(f(x, y)\) over \(D\) is \(\boxed{0.75 - \ln(2)}\).