Answer: (a) \(2x^2+5x-\frac12\ln(2x+3)+c\); (b) \(26-\frac12\ln\frac95\).

We start with the main method. Integrate term by term, remembering that an integral of the form \(1/(ax+b)\) gives a logarithm with factor \(1/a\).

(a) Integrate term by term:

\(\int 4x\,\mathrm dx=2x^2,\)

\(\int 5\,\mathrm dx=5x.\)

For the final term, use

\(\int\frac1{2x+3}\,\mathrm dx=\frac12\ln(2x+3).\)

Therefore

\(\int\left(4x+5-\frac1{2x+3}\right)\mathrm dx =2x^2+5x-\frac12\ln(2x+3)+c.\)

(b) Use the result from part (a):

\(\left[2x^2+5x-\frac12\ln(2x+3)\right]_1^3.\)

At \(x=3\), this is

\(18+15-\frac12\ln9=33-\frac12\ln9.\)

At \(x=1\), this is

\(2+5-\frac12\ln5=7-\frac12\ln5.\)

Subtracting gives

\(33-\frac12\ln9-\left(7-\frac12\ln5\right).\)

So the exact value is

\(26-\frac12\ln9+\frac12\ln5.\)

Hence

\(26-\frac12\ln\frac95.\)

This completes the solution and gives the required result.