已知 $y = (x + 2)\sqrt{x - 1}$，证明 $\frac{\text{d}y}{\text{d}x} = \frac{kx}{2\sqrt{x - 1}}$，其中 $k$ 为常数。 由此 已知 $y$ 以每秒 2 个单位的恒定速率变化，求当 $x = 2$ 时 $x$ 的变化率， 计算 $\int_{2}^{5} \frac{x}{\sqrt{x - 1}} \text{d}x$。

1. Apply the product rule to differentiate $y = (x + 2)(x - 1)^{\frac{1}{2}}$

Let $u = x + 2$ and $v = (x - 1)^{\frac{1}{2}}$.
$\frac{\text{d}u}{\text{d}x} = 1$
$\frac{\text{d}v}{\text{d}x} = \frac{1}{2}(x - 1)^{-\frac{1}{2}}(1) = \frac{1}{2\sqrt{x - 1}}$
$\frac{\text{d}y}{\text{d}x} = u\frac{\text{d}v}{\text{d}x} + v\frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}y}{\text{d}x} = (x + 2)\left(\frac{1}{2\sqrt{x - 1}}\right) + \sqrt{x - 1}(1)$

2. Combine the terms into a single fraction with a common denominator

$\frac{\text{d}y}{\text{d}x} = \frac{x + 2}{2\sqrt{x - 1}} + \frac{2(x - 1)}{2\sqrt{x - 1}}$
$\frac{\text{d}y}{\text{d}x} = \frac{x + 2 + 2x - 2}{2\sqrt{x - 1}}$

3. Simplify the numerator to reach the required form

$\frac{\text{d}y}{\text{d}x} = \frac{3x}{2\sqrt{x - 1}}$
Comparing with $\frac{kx}{2\sqrt{x - 1}}$, we have $k = 3$.

1. Determine the value of $k$ from part (i)

From part (i), differentiating $y = (x + 2)\sqrt{x - 1}$ using the product rule gives:
$\frac{\text{d}y}{\text{d}x} = (1)\sqrt{x - 1} + (x + 2)\left(\frac{1}{2\sqrt{x - 1}}\right)$
$\frac{\text{d}y}{\text{d}x} = \frac{2(x - 1) + (x + 2)}{2\sqrt{x - 1}} = \frac{3x}{2\sqrt{x - 1}}$
Thus, $k = 3$.

2. Evaluate $\frac{\text{d}y}{\text{d}x}$ at $x = 2$

Substitute $x = 2$ into $\frac{\text{d}y}{\text{d}x}$:
$\frac{\text{d}y}{\text{d}x} = \frac{3(2)}{2\sqrt{2 - 1}} = \frac{6}{2} = 3$

3. Use the chain rule to find the rate of change of $x$

Using the chain rule for connected rates of change, $\frac{\text{d}y}{\text{d}t} = \frac{\text{d}y}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}$.
Given $\frac{\text{d}y}{\text{d}t} = 2$:
$2 = 3 \times \frac{\text{d}x}{\text{d}t}$
$\frac{\text{d}x}{\text{d}t} = \frac{2}{3}$

1. Use the result from part (i) to set up the integral

From (i), we know that $\frac{\text{d}}{\text{d}x} [(x + 2)\sqrt{x - 1}] = \frac{3x}{2\sqrt{x - 1}}$.
Therefore, $\int \frac{3x}{2\sqrt{x - 1}} \text{d}x = (x + 2)\sqrt{x - 1} + c$.

2. Rearrange the integral to match the target expression

$\frac{3}{2} \int \frac{x}{\sqrt{x - 1}} \text{d}x = (x + 2)\sqrt{x - 1} + c$
$\int \frac{x}{\sqrt{x - 1}} \text{d}x = \frac{2}{3}(x + 2)\sqrt{x - 1} + c'$

3. Evaluate the definite integral using the limits 2 and 5

$\int_{2}^{5} \frac{x}{\sqrt{x - 1}} \text{d}x = \left[ \frac{2}{3}(x + 2)\sqrt{x - 1} \right]_{2}^{5}$
$= \frac{2}{3}(5 + 2)\sqrt{5 - 1} - \frac{2}{3}(2 + 2)\sqrt{2 - 1}$
$= \frac{2}{3}(7)(2) - \frac{2}{3}(4)(1)$
$= \frac{28}{3} - \frac{8}{3}$
$= \frac{20}{3} = 6\frac{2}{3}$