Answer: \(\ln4-\frac58\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

The intersection in the shaded region is found from

\(\frac{4}{2x+1}=3x+\frac12\).

Equivalently,

\(\frac{8}{2x+1}=6x+1\), so

\(8=(6x+1)(2x+1)\).

This gives

\(12x^2+8x-7=0\), and the positive solution is \(x=\frac12\).

The shaded area is

\(\int_0^{1/2}\left(\frac{4}{2x+1}-\left(3x+\frac12\right)\right)\,dx\).

Now

\(\int_0^{1/2}\frac{4}{2x+1}\,dx=\left[2\ln(2x+1)\right]_0^{1/2}=2\ln2=\ln4\).

Also

\(\int_0^{1/2}\left(3x+\frac12\right)\,dx=\left[\frac32x^2+\frac12x\right]_0^{1/2}=\frac58\).

Therefore the required area is

\(\ln4-\frac58\).

This gives the required answer: \(\ln4-\frac58\).