The diagram shows part of the curve $y = \frac{4x + 1}{x - 2}$.
A line intersects the curve at points $P(11, 5)$ and $Q(3, 13)$. By expressing $\frac{4x + 1}{x - 2}$ in the form $a + \frac{b}{x - 2}$, where $a$ and $b$ are constants, find, showing full working, the area of the shaded region.

1. Express the rational function in the required form

$\frac{4x + 1}{x - 2} = \frac{4(x - 2) + 8 + 1}{x - 2} = 4 + \frac{9}{x - 2}$

2. Find the equation of the line $QP$

Gradient $m = \frac{13 - 5}{3 - 11} = \frac{8}{-8} = -1$
Equation of line: $y - 5 = -1(x - 11) \implies y = -x + 16$

3. Set up the expression for the shaded area

Area = $\int_{3}^{11} (-x + 16) \, dx - \int_{3}^{11} \left(4 + \frac{9}{x - 2}\right) \, dx$
Alternatively, Area of trapezium under $QP$ = $\frac{1}{2}(13 + 5)(11 - 3) = 72$

4. Integrate the curve equation

$\int \left(4 + \frac{9}{x - 2}\right) \, dx = 4x + 9\ln(x - 2)$

5. Evaluate the definite integral for the area under the curve

Area under curve = $\left[ 4x + 9\ln(x - 2) \right]_{3}^{11}$
$= (44 + 9\ln 9) - (12 + 9\ln 1)$
$= 32 + 9\ln 9$

6. Calculate the final shaded area

Shaded Area = $72 - (32 + 9\ln 9) = 40 - 9\ln 9$