Answer: \(y_A=\frac52\), \(y_B=1\), and shaded area \(=\frac{11}{4}-\ln\frac{27}{2}\). Hence \(a=11\), \(b=27\).

At \(A\), \(x=1\), so

\(y=\frac{3}{2}-\frac{1-2}{1}=\frac32+1=\frac52.\)

At \(B\), \(x=2\), so

\(y=\frac{3}{3}-\frac{2-2}{2}=1.\)

The area of the trapezium under the line \(AB\) between \(x=1\) and \(x=2\) is

\(\frac12\left(\frac52+1\right)(1)=\frac74.\)

The area under the curve is

\(\int_1^2\left(\frac{3}{x+1}-\frac{x-2}{x}\right)\,\mathrm{d}x.\)

Since

\(\frac{3}{x+1}-\frac{x-2}{x}=\frac{3}{x+1}-1+\frac{2}{x},\)

an antiderivative is

\(3\ln(x+1)-x+2\ln x.\)

Thus the area under the curve from \(1\) to \(2\) is

\((3\ln3-2+2\ln2)-(3\ln2-1)=\ln\frac{27}{2}-1.\)

Therefore the shaded area is

\(\frac74-\left(\ln\frac{27}{2}-1\right)=\frac{11}{4}-\ln\frac{27}{2}.\)

So \(a=11\) and \(b=27\).