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 compute the integral, we need:

$\int_{0}^{2} \frac{1}{(t + 1)(t + 2)} dt$

Using partial fraction decomposition, we can write:

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

Multiplying through by the denominator $(t + 1)(t + 2)$ gives:

$1 = A(t + 2) + B(t + 1)$

Setting up equations by choosing convenient values:

1. Let $t = -1 \Rightarrow 1 = B(1) \Rightarrow B = 1$.
2. Let $t = -2 \Rightarrow 1 = A(-1) \Rightarrow A = -1$.

Hence, we have:

$\frac{1}{(t + 1)(t + 2)} = \frac{-1}{t + 1} + \frac{1}{t + 2}$

Now substituting back into the integral gives:

$\int_{0}^{2} \left( \frac{-1}{t + 1} + \frac{1}{t + 2} \right) dt$

Calculating these integrals separately:

1. For $\int_{0}^{2} \frac{-1}{t + 1} dt = -\ln(t + 1) \Big|_{0}^{2} = -[\ln(3) - \ln(1)] = -\ln(3)$.
2. For $\int_{0}^{2} \frac{1}{t + 2} dt = \ln(t + 2) \Big|_{0}^{2} = [\ln(4) - \ln(2)] = \ln(2)$.

Therefore, combining these results:

$\text{Area}(R) = \ln(2) - \ln(3) = \ln\left(\frac{2}{3}\right).$

We start with our parametric equations:

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

From the expression for $x$, we can isolate $t$:

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

Substituting this expression for $t$ back into the equation for $y$ gives:

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

Thus, the Cartesian equation is:

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

For the equation $y = \frac{1}{e^x - 1}$ to be defined, the denominator must be non-zero:

$e^x - 1 \neq 0 \Rightarrow e^x \neq 1 \Rightarrow x \neq 0.$

Moreover, since $t > -1$, this corresponds to:

$x = \text{ln}(t + 2) > \text{ln}(1) = 0.$

Therefore, the domain of values for x is:

$x > 0.$