已知 $y = (x-3)\sqrt{x+1}$。以 $\dfrac{ax-b}{2\sqrt{x+1}}$ 的形式求 $\dfrac{dy}{dx}$，其中 $a$ 和 $b$ 为整数。 由此求 $\displaystyle\int_0^3 \dfrac{9x}{\sqrt{x+1}}\,dx$ 的值。

1. Apply the product rule to differentiate the function

Let $y = (x-3)(x+1)^{\frac{1}{2}}$.
Using the product rule, $\dfrac{dy}{dx} = u \dfrac{dv}{dx} + v \dfrac{du}{dx}$:
$\dfrac{dy}{dx} = (x-3) \cdot \dfrac{1}{2}(x+1)^{-\frac{1}{2}} + (x+1)^{\frac{1}{2}} \cdot (1)$

2. Express the derivative as a single algebraic fraction

$\dfrac{dy}{dx} = \dfrac{x-3}{2\sqrt{x+1}} + \sqrt{x+1}$
$= \dfrac{x-3 + 2(x+1)}{2\sqrt{x+1}}$

3. Simplify the numerator to match the required form

$= \dfrac{x - 3 + 2x + 2}{2\sqrt{x+1}}$
$= \dfrac{3x - 1}{2\sqrt{x+1}}$

1. Find the derivative from part (a) to use for integration

From part (a), let $y = (x-3)\sqrt{x+1}$.
Using the product rule:
$\frac{dy}{dx} = (1)\sqrt{x+1} + (x-3)\left(\frac{1}{2\sqrt{x+1}}\right)$
$\frac{dy}{dx} = \frac{2(x+1) + x - 3}{2\sqrt{x+1}} = \frac{3x-1}{2\sqrt{x+1}}$

2. Express the required integrand in terms of the derivative from part (a)

We want to integrate $\frac{9x}{\sqrt{x+1}}$. Notice that:
$\frac{9x}{\sqrt{x+1}} = 6 \left( \frac{3x}{2\sqrt{x+1}} \right)$
We can rewrite $\frac{3x}{2\sqrt{x+1}}$ using the derivative:
$\frac{3x}{2\sqrt{x+1}} = \frac{3x-1}{2\sqrt{x+1}} + \frac{1}{2\sqrt{x+1}}$
Therefore:
$\frac{9x}{\sqrt{x+1}} = 6 \left( \frac{d}{dx}\left[(x-3)\sqrt{x+1}\right] + \frac{1}{2}(x+1)^{-\frac{1}{2}} \right)$

3. Integrate the expression using the reverse of differentiation

$\int_0^3 \frac{9x}{\sqrt{x+1}} \,dx = 6 \int_0^3 \left( \frac{d}{dx}\left[(x-3)\sqrt{x+1}\right] + \frac{1}{2}(x+1)^{-\frac{1}{2}} \right) \,dx$
$= 6 \left[ (x-3)\sqrt{x+1} + \frac{\frac{1}{2}(x+1)^{\frac{1}{2}}}{\frac{1}{2}} \right]_0^3$
$= 6 \left[ (x-3)\sqrt{x+1} + \sqrt{x+1} \right]_0^3$
$= 6 \left[ (x-2)\sqrt{x+1} \right]_0^3$

4. Evaluate the definite integral

$= 6 \left( \left[ (3-2)\sqrt{3+1} \right] - \left[ (0-2)\sqrt{0+1} \right] \right)$
$= 6 \left( (1)(2) - (-2)(1) \right)$
$= 6 (2 + 2)$
$= 24$