Figure 1 shows a sketch of part of the curve with equation $y = 4x - xe^{-x}, \, x > 0$ The curve meets the x-axis at the origin $O$ and cuts the x-axis at the point $A$ - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 4

For this integration, we can use integration by parts: Let $u = e^{-\frac{x}{2}}, \\ dv = \frac{1}{x} dx.$

Then, calculate: $du = -\frac{1}{2} e^{-\frac{x}{2}} \, dx, \\ v = \ln|x|.$

Using the integration by parts formula, we have: $\int u \, dv = uv - \int v \, du.$

Thus, it becomes: $= e^{-\frac{x}{2}} \ln|x| + \frac{1}{2} \int e^{-\frac{x}{2}} dx.$

The last integral can be integrated directly to yield: $= -2 e^{-\frac{x}{2}} + C.$

This integral evaluates to a function that can be simplified further.

To find the area of the region $R$, we will integrate between the points $O$ and $A$:

$\int_0^a (4x - xe^{-x}) \, dx$

Computing the integral: $= \left[ 2x^2 - 2x^2 e^{-x} \right]_0^a \\ = 2a^2 - 2(a^2e^{-a}) = 2a^2(1 - e^{-a}).$

Substituting $a = -2\ln(2)$ into this final result will yield the exact area in terms of $ext{ln}(2)$. After calculating, you will find:

$\text{Area } = 4\ln(2) - 8.$

Therefore, the area of region R will be expressed clearly in terms of $\ln(2)$.