The diagram below shows the graph of $y = \frac{1}{2x+1}$ and the tangent to the curve where the curve intersects the $y$-axis.

1. Find the coordinates of the point where the curve intersects the $y$-axis

When $x = 0$,
$y = \frac{1}{2(0)+1} = 1$
The point of intersection is $(0, 1)$.

2. Differentiate the equation of the curve to find the gradient function

$y = (2x+1)^{-1}$
$\frac{dy}{dx} = -1(2x+1)^{-2}(2)$
$\frac{dy}{dx} = -\frac{2}{(2x+1)^2}$

3. Find the gradient of the tangent at $x = 0$ and form the equation of the tangent

At $x = 0$, gradient $m = -\frac{2}{(2(0)+1)^2} = -2$
Equation of tangent:
$y - 1 = -2(x - 0)$
$y = -2x + 1$

1. Find the equation of the tangent and its x-intercept (from part a)

$\frac{dy}{dx} = -2(2x+1)^{-2}$
At $x = 0$, $\frac{dy}{dx} = -2$.
The equation of the tangent at $(0, 1)$ is $y - 1 = -2(x - 0) \implies y = -2x + 1$.
When $y = 0$, $-2x + 1 = 0 \implies x = 0.5$.
The tangent intersects the $x$-axis at $x = 0.5$.

2. Set up the expression for the shaded area

The shaded area is the area under the curve from $x = 0$ to $x = 1$ minus the area of the triangle formed by the tangent line and the axes.
$\text{Area} = \int_{0}^{1} \frac{1}{2x+1} \, dx - \frac{1}{2}(\text{base})(\text{height})$

3. Evaluate the integral of the curve

$\int_{0}^{1} \frac{1}{2x+1} \, dx = \left[ \frac{1}{2} \ln(2x+1) \right]_{0}^{1}$
$= \frac{1}{2}\ln(3) - \frac{1}{2}\ln(1)$
$= \frac{1}{2}\ln 3$

4. Calculate the final area

$\text{Area of triangle} = \frac{1}{2}(0.5)(1) = 0.25 = \frac{1}{4}$
$\text{Shaded Area} = \frac{1}{2}\ln 3 - \frac{1}{4}$