Answer: \(\ln 5-\dfrac67\).

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 \(1/a\).

Integrate each term separately:

\(\int \frac{2}{2x-3}\,dx=\ln(2x-3),\)

and

\(\int -\frac{3}{(3x-5)^2}\,dx=\frac{1}{3x-5}.\)

So an antiderivative is

\(\ln(2x-3)+\frac{1}{3x-5}.\)

Now apply the limits \(2\) and \(4\):

\(\left[\ln(2x-3)+\frac{1}{3x-5}\right]_2^4\)

\(=\left(\ln5+\frac17\right)-\left(\ln1+1\right).\)

Since \(\ln1=0\), the value is

\(\ln5+\frac17-1=\ln5-\frac67.\)

This completes the solution and gives the required result.