Answer: (a) \(\frac35(\sqrt[3]{18}-2)\); (b) \(2+\ln4\).

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:

\(\int(5x-2)^{-2/3}\,\mathrm dx=\frac35(5x-2)^{1/3}.\)

Therefore

\(\int_2^4(5x-2)^{-2/3}\,\mathrm dx =\left[\frac35(5x-2)^{1/3}\right]_2^4.\)

So

\(\frac35(18^{1/3}-8^{1/3})=\frac35(\sqrt[3]{18}-2).\)

(b) Integrate term by term:

\(\int\frac4{2x+1}\,\mathrm dx=2\ln(2x+1),\)

and

\(\int\frac8{(2x+1)^2}\,\mathrm dx=-\frac4{2x+1}.\)

Thus

\(\int_0^{1/2}\left(\frac4{2x+1}+\frac8{(2x+1)^2}\right)\mathrm dx =\left[2\ln(2x+1)-\frac4{2x+1}\right]_0^{1/2}.\)

At \(x=\frac12\), this is \(2\ln2-2\). At \(x=0\), this is \(-4\).

Therefore the value is

\((2\ln2-2)-(-4)=2\ln2+2.\)

Since \(2\ln2=\ln4\), the answer is

\(2+\ln4.\)

This completes the solution and gives the required result.