Use the substitution $u = ext{log}_e x$ to evaluate \[ \int_{e}^{e^2} \frac{1}{x \left(\text{log}_e x\right)^2} \, dx - HSC - SSCE Mathematics Extension 1 - Question 2 - 2008 - Paper 1

The acceleration is given by $\bar{x} = \frac{d}{dx} \left(\frac{1}{2} v^2\right)$. We have:

[\frac{d}{dx} \left(\frac{1}{2} v^2\right) = v \cdot \frac{dv}{dx}.]

Given that the acceleration is $\bar{x} = \frac{1}{2} v^2 + 4$, we can set:

[v \cdot \frac{dv}{dx} = \frac{1}{2} v^2 + 4.]

Separating variables and integrating leads us to find the value of $v$ at $x = 2$. Initially at rest, we can calculate using initial conditions. After solving, we'll find the speed at $x = 2$.

Considering $p(x)$, we know the roots are -2, 3, and $\alpha$. Using Vieta's relations, we have:

[r_1 + r_2 + r_3 = -\frac{16}{a}]

Substituting the known roots:

[-2 + 3 + \alpha = -\frac{16}{a}]

This simplifies to:

[1 + \alpha = -\frac{16}{a} \rightarrow \alpha = -\frac{16}{a} - 1.]

Additionally, we can consider the product of roots, giving us another equation to solve for $\alpha$. By substituting the values from the equation, we arrive at $\alpha$.

We will apply Newton's method to the function $f(x) = \tan x - \log_e x$. To find a zero near $x = 4$, we first compute:

[f'(x) = \sec^2 x - \frac{1}{x}. ]

Evaluating $f(4)$ and $f'(4)$:

1. Calculate $f(4)$ and $f'(4)$.
2. Apply Newton's formula:

[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.]

Using one iteration, compute $x_5$ to two decimal places.