The curve C has parametric equations $x = ext{ln}(t + 2),$ $y = rac{1}{(t + 1)},$ $t > -1.$ The finite region R between the curve C and the x-axis, bounded by the lines with equations $x = ext{ln} 2$ and $x = ext{ln} 4$, is shown shaded in Figure 3 - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 8

To find the exact area, we can compute the integral:

$A = \int_{0}^{2} \frac{1}{(t + 1)(t + 2)} \, dt.$

We can perform partial fraction decomposition:

$\frac{1}{(t + 1)(t + 2)} = \frac{A}{t + 1} + \frac{B}{t + 2}.$

By solving:

1. $A(t + 2) + B(t + 1) = 1$
2. Setting $t = -1$, we find $A = 1$.
3. Setting $t = -2$, we find $B = -1$.

Thus:

$A = \int_{0}^{2} \left( \frac{1}{t + 1} - \frac{1}{t + 2} \right) \, dt = \left[\ln(t + 1) - \ln(t + 2) \right]_{0}^{2}.$

Evaluating this gives:

$\left( \ln(3) - \ln(4) \right) - \left( \ln(1) - \ln(2) \right) = \ln(3) - \ln(4) + \ln(2) = \ln(\frac{3}{2}).$

We start with the parametric equations:

1. $x = \text{ln}(t + 2)$
2. $y = \frac{1}{t + 1}$.

To eliminate $t$, we solve the first equation for $t$:

$t + 2 = e^x \implies t = e^x - 2.$

Now substitute $t$ from the first equation into the second:

$y = \frac{1}{(e^x - 2) + 1} = \frac{1}{e^x - 1}.$

Thus, the cartesian equation of the curve C is:

$y = \frac{1}{e^x - 1}.$

For the curve $y = \frac{1}{e^x - 1}$ to be defined, the denominator cannot be zero:

1. Setting $e^x - 1 > 0 \implies e^x > 1 \implies x > 0.$

Thus, the domain of values for $x$ is:

$x > 0.$