Answer: \(\frac{13}{2}+4\ln\frac32\).

We start with the main method. Simplify the integrand before integrating, then substitute the upper and lower limits carefully.

First expand the integrand:

\(\frac{(x+2)^2}{x}=\frac{x^2+4x+4}{x}=x+4+\frac4x.\)

So

\(\int_2^3\frac{(x+2)^2}{x}\,\mathrm dx =\int_2^3\left(x+4+\frac4x\right)\mathrm dx.\)

Integrating term by term gives

\(\left[\frac{x^2}{2}+4x+4\ln x\right]_2^3.\)

At \(x=3\), this is

\(\frac92+12+4\ln3.\)

At \(x=2\), this is

\(2+8+4\ln2.\)

Therefore the exact value is

\(\left(\frac92+12+4\ln3\right)-\left(2+8+4\ln2\right).\)

So

\(\int_2^3\frac{(x+2)^2}{x}\,\mathrm dx =\frac{13}{2}+4\ln\frac32.\)

This completes the solution and gives the required result.