Answer: \(A=\left(\frac32,4\right)\), and the shaded area is \(5\ln\left(\frac52\right)-\frac34\).

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.

At the point of intersection,

\(\frac{5}{x+1}+2=2x+1.\)

So

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

Multiplying by \(x+1\),

\(5=(2x-1)(x+1).\)

Hence

\(2x^2+x-1=5,\)

so

\(2x^2+x-6=0.\)

Factorise:

\((2x-3)(x+2)=0.\)

The positive intersection shown is \(x=\frac32\). Then

\(y=2\left(\frac32\right)+1=4.\)

So \(A=\left(\frac32,4\right)\).

The shaded region is between \(x=0\) and \(x=\frac32\), with the curve above the line. Its area is

\(\int_0^{3/2}\left(\frac{5}{x+1}+2-(2x+1)\right)\,\mathrm{d}x.\)

This simplifies to

\(\int_0^{3/2}\left(\frac{5}{x+1}+1-2x\right)\,\mathrm{d}x.\)

An antiderivative is

\(5\ln(x+1)+x-x^2.\)

Therefore the area is

\(\left[5\ln(x+1)+x-x^2\right]_0^{3/2}.\)

Substituting the limits gives

\(5\ln\left(\frac52\right)+\frac32-\frac94=5\ln\left(\frac52\right)-\frac34.\)

This gives the required answer: \(A=\left(\frac32,4\right)\), and the shaded area is \(5\ln\left(\frac52\right)-\frac34\).