Example 4. Calculate $\int_{0}^{5} x \sqrt{x+4} d x$.
Solution. Introduce a new variable by the formula $\sqrt{x+4}=t$. Determine $x$ and $d x$. Squaring both sides of the equation $\sqrt{x+4}=t$, we get $x+4=t^{2}$, from which $x=t^{2}-4$ and $d x=2 t d t$. Find the new limits of integration. Substituting the old limits into the formula $\sqrt{x+4}=t$, we get: $\sqrt{0+4}=t$, from which $t=2$ and $\alpha=2 ; \sqrt{5+4}=t$, from which $t=3$, $\beta=3$.

Thus,

$$
\begin{aligned}
& \int_{0}^{5} x \sqrt{x+4} d x=\int_{2}^{3}\left(t^{2}-4\right) t 2 t d t=\int_{2}^{3}\left(2 t^{4}-8 t^{2}\right) d t=2 \int_{2}^{3} t^{4} d t- \\
&-8 \int_{2}^{3} t^{2} d t=\left.2 \cdot \frac{t^{5}}{5}\right|_{2} ^{3}-\left.8 \cdot \frac{t^{3}}{3}\right|_{2} ^{3}=\frac{2}{5}\left(3^{5}-2^{5}\right)-\frac{8}{3}\left(3^{3}-2^{3}\right)= \\
&=\frac{2}{5} \cdot 211-\frac{8}{3} \cdot 19=\frac{506}{15}=33 \frac{11}{15}
\end{aligned}
$$