Answer: \(\dfrac{18}{25}+\dfrac23\ln\left(\dfrac{24\sqrt3}{5}\right)\).

(a) Use the common denominator \((x+1)(3x+10)\):

\(\frac{1}{x+1}+\frac{2}{3x+10} =\frac{3x+10+2(x+1)}{(x+1)(3x+10)}.\)

Simplifying,

\(\frac{3x+10+2x+2}{3x^2+13x+10} =\frac{5x+12}{3x^2+13x+10}.\)

(b) Since

\(y=\frac{1}{x+1}+\frac{2}{3x+10},\)

the point \(P\) is found from \(x=0\):

\(y=1+\frac{2}{10}=\frac65.\)

So

\(P=\left(0,\frac65\right).\)

The straight line has gradient \(1\), so its equation is

\(y=x+\frac65.\)

It meets the \(x\)-axis when \(y=0\), so

\(x=-\frac65.\)

Thus

\(Q=\left(-\frac65,0\right).\)

The triangular part of the shaded area is

\(\frac12\cdot\frac65\cdot\frac65=\frac{18}{25}.\)

The curved part of the shaded area is

\(\int_0^2\left(\frac{1}{x+1}+\frac{2}{3x+10}\right)\,dx.\)

Integrating gives

\(\left[\ln(x+1)+\frac23\ln(3x+10)\right]_0^2.\)

Using the limits,

\(\ln3+\frac23\ln16-\frac23\ln10.\)

Since

\(\ln3=\frac23\ln(3\sqrt3),\)

the integral becomes

\(\frac23\ln(3\sqrt3)+\frac23\ln\left(\frac{16}{10}\right) =\frac23\ln\left(\frac{24\sqrt3}{5}\right).\)

Therefore the total shaded area is

\(\frac{18}{25}+\frac23\ln\left(\frac{24\sqrt3}{5}\right).\)