The diagram shows part of the curve $y = \frac{8}{2x+3}$. The tangent to the curve at the point $S\left(\frac{3}{2}, \frac{4}{3}\right)$ intersects the $y$-axis at $R$.

1. Find the gradient of the tangent at $S$ to determine the $y$-coordinate of $R$ (from part a).

$\frac{dy}{dx} = -8(2x+3)^{-2}(2) = -\frac{16}{(2x+3)^2}$
At $x = \frac{3}{2}$, $\frac{dy}{dx} = -\frac{16}{(3+3)^2} = -\frac{16}{36} = -\frac{4}{9}$
Equation of tangent: $y - \frac{4}{3} = -\frac{4}{9}\left(x - \frac{3}{2}\right)$
$y = -\frac{4}{9}x + \frac{2}{3} + \frac{4}{3} = -\frac{4}{9}x + 2$
Thus, the $y$-coordinate of $R$ is $2$.

2. Calculate the area under the tangent line (a trapezium) from $x = 0$ to $x = \frac{3}{2}$.

Area of trapezium = $\frac{1}{2} \times \left(2 + \frac{4}{3}\right) \times \frac{3}{2}$
$= \frac{1}{2} \times \frac{10}{3} \times \frac{3}{2} = \frac{5}{2}$

3. Calculate the area under the curve from $x = 0$ to $x = \frac{3}{2}$.

Area under curve = $\int_{0}^{\frac{3}{2}} \frac{8}{2x+3} \, dx$
$= \left[ \frac{8}{2} \ln(2x+3) \right]_{0}^{\frac{3}{2}}$
$= \left[ 4 \ln(2x+3) \right]_{0}^{\frac{3}{2}}$
$= 4 \ln\left(2\left(\frac{3}{2}\right)+3\right) - 4 \ln(2(0)+3)$
$= 4 \ln 6 - 4 \ln 3$

4. Find the exact area of the shaded region by subtracting the area of the trapezium from the area under the curve.

Shaded area = $(4 \ln 6 - 4 \ln 3) - \frac{5}{2}$
$= 4 \ln\left(\frac{6}{3}\right) - \frac{5}{2}$
$= 4 \ln 2 - \frac{5}{2}$
$= \ln(2^4) - \frac{5}{2}$
$= \ln 16 - \frac{5}{2}$