Figure 1 shows a sketch of part of the curve with equation $y = \frac{6}{e^x + 2}$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 5

The area $A_R$ using the trapezium rule is given by: $A_R \approx \frac{1}{2} (y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + y_5) \Delta x$
where $\Delta x = 0.2$ and the corresponding $y$ values are:
$y_0 = 2, \; y_1 = 1.86522, \; y_2 = 1.71830, \; y_3 = 1.56981, \; y_4 = 1.41994, \; y_5 = 1.27165$
Calculating this yields:

$A_R \approx \frac{1}{2} (2 + 2(1.86522 + 1.71830 + 1.56981 + 1.41994) + 1.27165) \cdot 0.2$
This gives: $A_R \approx 0.41283$
Consequently, the estimated area of region $R$ rounded to 4 decimal places is approximately $1.6413$.

Starting with: $x = \ln(u) \quad \Rightarrow \quad dx = \frac{1}{u} du$
The bounds change as $x$ goes from $0$ to $1$: when $x = 0$, $u = e^0 = 1$, and when $x = 1$, $u = e^1 = e$. Substituting in:

$\int_{1}^{e} \frac{6}{e^x + 2} dx \Rightarrow \int_{1}^{e} \frac{6}{u + 2} \cdot \frac{1}{u} du$
This yields: $\int_{1}^{e} \frac{6}{u(u + 2)} du$
Therefore, the area of $R$ can be expressed as
$A_R = \int_1^e \frac{6}{u(u + 2)} du$.

To evaluate the integral: $A_R = \int_1^e \frac{6}{u(u + 2)} du$, we can use partial fractions: $\frac{6}{u(u + 2)} = \frac{A}{u} + \frac{B}{u + 2}$. Solving for constants $A$ and $B$, we find: $A = 3, \; B = -3$, leading to: $A_R = \int_1^e \left( \frac{3}{u} - \frac{3}{u + 2} \right) du$. Evaluating this gives: $A_R = 3(\ln|u| - \ln|u + 2|) \Big|_1^e \rightarrow 3(\ln(e) - \ln(e + 2) - (\ln(1) - \ln(3)))$. This yields: $A_R = 3(1 - \ln(e + 2) + \ln(3)) = 3 - 3\ln(5)$. Ultimately, the exact area of $R$ is: $A_R = 3 - 3\ln(5)$.