Example 5. Calculate $\int_{0}^{4} \frac{d x}{\sqrt{x}+1}$.
Solution. Let $\sqrt{x}=t$, then $x=t^{2}$ and $d x=2 t d t$. Substituting the old limits of integration into the formula $\sqrt{x}=t$, we get $\alpha=0$, $\beta=2$.

Therefore,

$$
\begin{gathered}
\int_{0}^{4} \frac{d x}{\sqrt{x}+1}=\int_{0}^{2} \frac{2 t d t}{t+1}=2 \int_{0}^{2} \frac{(t+1)-1}{t+1} d t=2 \int_{0}^{2} d t-2 \int_{0}^{2} \frac{d(t+1)}{t+1}= \\
=\left.2 t\right|_{0} ^{2}-\left.2 \ln |t+1|\right|_{0} ^{2}=2(2-0)-2[\ln (2+1)-\ln (0+1)]= \\
=4-2 \ln 3 \approx 1.803
\end{gathered}
$$