Example 3.13. Using the Newton-Leibniz formula, compute the definite integral

$$
\int_{3}^{8} \frac{x}{\sqrt{1+x}} d x
$$
Solution. To get rid of the irrationality, we make the substitution $1+x=t^{2}$. Then

$$
\begin{aligned}
& \int_{3}^{8} \frac{x d x}{\sqrt{1+x}}=\left|\begin{array}{l}
1+x=t^{2} \\
d x=2 t d t \\
x=3 \rightarrow t=2 \\
x=8 \rightarrow t=3
\end{array}\right|=\int_{2}^{3} \frac{\left(t^{2}-1\right) 2 t d t}{t}=2 \int_{2}^{3}\left(t^{2}-1\right) d t= \\
& =2 \frac{t^{3}}{3}-\left.t\right|_{2} ^{3}=2 \frac{27}{3}-3-\frac{8}{3}+2=\frac{32}{3}
\end{aligned}
$$

The formula for integration by parts in a definite integral:

$$
\int_{a}^{b} u(x) d v(x)=\left.u(x) v(x)\right|_{a} ^{b}-\int_{a}^{b} v(x) d u(x)
$$

where $\left.u(x) v(x)\right|_{a} ^{b}$ is calculated according to formula (3.6).