Example 6 Calculate the integral $\int_{5}^{12} \frac{\sqrt{x+4}}{x} d x$ and find the integral mean value of the integrand $f(x)=\frac{\sqrt{x+4}}{x}$ over the interval of integration $[5 ; 12]$.
Solution. We integrate by substitution. In this process, we transition to another definite integral with a different integrand and different limits.

Let $\sqrt{x+4}=u$. Then $x+4=u^{2}, x=u^{2}-4, d x=2 u d u$. Therefore, if $x=5$, then $u=3$, and if $x=12$, then $u=4$. We obtain

$$
\begin{aligned}
\int_{5}^{12} \frac{\sqrt{x+4}}{x} d x=\int_{3}^{4} \frac{u \cdot 2 u d u}{u^{2}-4} & =2 \int_{3}^{4} \frac{u^{2}-4+4}{u^{2}-4} d u= \\
& =\left.2\left(u+\ln \frac{u-2}{u+2}\right)\right|_{3} ^{4}=2\left(1+\ln \frac{5}{3}\right)=2 \ln \frac{5 e}{3}
\end{aligned}
$$

The length of the integration interval is $12-5=7$, and the integral mean value of the function is $\frac{2}{7} \ln \frac{5 e}{3}$.

Answer. $2 \ln \frac{5 e}{3} ; \frac{2}{7} \ln \frac{5 e}{3}$