Answer: \(\dfrac94+2\ln2\).

First expand the numerator:

\((x+1)^2=x^2+2x+1.\)

Therefore

\(\frac{(x+1)^2}{x^2}=1+\frac2x+\frac1{x^2}.\)

So

\(\int_2^4\frac{(x+1)^2}{x^2}\,dx =\int_2^4\left(1+\frac2x+\frac1{x^2}\right)\,dx.\)

An antiderivative is

\(x+2\ln x-\frac1x.\)

Evaluate between \(2\) and \(4\):

\(\left[x+2\ln x-\frac1x\right]_2^4.\)

This is

\(\left(4+2\ln4-\frac14\right)-\left(2+2\ln2-\frac12\right).\)

Since \(\ln4=2\ln2\), this becomes

\(2+2\ln2+\frac14.\)

Hence the exact value is

\(\frac94+2\ln2.\)