一曲线的方程为 $y = 2x\sqrt{2x+1}$。

1. Differentiate the function using the product rule and chain rule.

$\frac{dy}{dx} = 2(2x+1)^{\frac{1}{2}} + 2x \left[ \frac{1}{2}(2x+1)^{-\frac{1}{2}}(2) \right]$
$= 2(2x+1)^{\frac{1}{2}} + 2x(2x+1)^{-\frac{1}{2}}$

2. Express the derivative as a single fraction to match the required form.

$\frac{dy}{dx} = \frac{2(2x+1)}{\sqrt{2x+1}} + \frac{2x}{\sqrt{2x+1}}$
$= \frac{4x + 2 + 2x}{\sqrt{2x+1}}$
$= \frac{6x+2}{\sqrt{2x+1}}$
(where $a=6$ and $b=2$)

1. Find the derivative from part (a) to determine the constants $a$ and $b$.

Let $y = 2x\sqrt{2x+1}$.
Using the product rule and chain rule:
$\frac{dy}{dx} = 2\sqrt{2x+1} + 2x \left( \frac{1}{2}(2x+1)^{-\frac{1}{2}}(2) \right)$
$= 2\sqrt{2x+1} + \frac{2x}{\sqrt{2x+1}}$
$= \frac{2(2x+1) + 2x}{\sqrt{2x+1}}$
$= \frac{6x+2}{\sqrt{2x+1}}$
$= \frac{2(3x+1)}{\sqrt{2x+1}}$

2. Use the reverse process of differentiation to evaluate the integral.

Since $\frac{d}{dx}(2x\sqrt{2x+1}) = 2 \left( \frac{3x+1}{\sqrt{2x+1}} \right)$,
$\int \frac{3x+1}{\sqrt{2x+1}} \, dx = \frac{1}{2} \int \frac{2(3x+1)}{\sqrt{2x+1}} \, dx$
$= \frac{1}{2} (2x\sqrt{2x+1}) + c$
$= x\sqrt{2x+1} + c$