29.8. Compute $\int \frac{d x}{1+\sqrt{x}}$.
29.8. Let $x=y^{2}$. Then, according to problem 29.5

$$
\begin{aligned}
& \int \frac{d x}{1+\sqrt{x}}=\int \frac{2 y d y}{1+y}=2 \int\left(1-\frac{1}{1+y}\right) d y= \\
& \quad=2 y-2 \ln (1+y)+C=2 \sqrt{x}-2 \ln (1+\sqrt{x})+C
\end{aligned}
$$