The equation of a curve is $y = x\sqrt{x+1}$.

1. Apply the product rule and chain rule to differentiate the expression.

Let $y = x(x+1)^{\frac{1}{2}}$.
Using the product rule:
$\frac{dy}{dx} = (1)(x+1)^{\frac{1}{2}} + x\left[\frac{1}{2}(x+1)^{-\frac{1}{2}}(1)\right]$

2. Express the derivative with positive indices.

$\frac{dy}{dx} = \sqrt{x+1} + \frac{x}{2\sqrt{x+1}}$

3. Combine the terms into a single fraction to obtain the required result.

$\frac{dy}{dx} = \frac{2(x+1)}{2\sqrt{x+1}} + \frac{x}{2\sqrt{x+1}}$
$= \frac{2x + 2 + x}{2\sqrt{x+1}}$
$= \frac{3x+2}{2\sqrt{x+1}} \text{ (Shown)}$

1. State the chain rule for connected rates of change

By the chain rule,
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$

2. Evaluate the derivative at $x=5$ and calculate the rate of change of $y$

From part (a), $\frac{dy}{dx} = \frac{3x+2}{2\sqrt{x+1}}$.
When $x = 5$,
$\frac{dy}{dx} = \frac{3(5)+2}{2\sqrt{5+1}} = \frac{17}{2\sqrt{6}}$
Given $\frac{dx}{dt} = 0.2$,
$\frac{dy}{dt} = \frac{17}{2\sqrt{6}} \times 0.2 = \frac{17}{10\sqrt{6}} = \frac{17\sqrt{6}}{60} \approx 0.694 \text{ units/s}$

1. Use the result from part (a) to set up an equation involving definite integrals.

From part (a), we know that $\frac{d}{dx}\left(x\sqrt{x+1}\right) = \frac{3x+2}{2\sqrt{x+1}}$.
Integrating both sides with respect to $x$ from $-1$ to $3$:
$\int_{-1}^{3} \frac{3x+2}{2\sqrt{x+1}} \,dx = \left[ x\sqrt{x+1} \right]_{-1}^{3}$

2. Split the integrand to isolate the required integral.

We can split the fraction:
$\int_{-1}^{3} \left( \frac{3x}{2\sqrt{x+1}} + \frac{2}{2\sqrt{x+1}} \right) \,dx = \left[ x\sqrt{x+1} \right]_{-1}^{3}$
$3 \int_{-1}^{3} \frac{x}{2\sqrt{x+1}} \,dx + \int_{-1}^{3} (x+1)^{-\frac{1}{2}} \,dx = \left[ x\sqrt{x+1} \right]_{-1}^{3}$

3. Evaluate the known integral and the right-hand side.

Evaluate the right-hand side:
$\left[ x\sqrt{x+1} \right]_{-1}^{3} = 3\sqrt{3+1} - (-1)\sqrt{-1+1} = 3(2) - 0 = 6$

Evaluate the remaining integral on the left:
$\int_{-1}^{3} (x+1)^{-\frac{1}{2}} \,dx = \left[ \frac{(x+1)^{\frac{1}{2}}}{\frac{1}{2}} \right]_{-1}^{3} = \left[ 2\sqrt{x+1} \right]_{-1}^{3}$
$= 2\sqrt{4} - 2\sqrt{0} = 4$

4. Solve for the required integral.

Substitute these values back into the equation:
$3 \int_{-1}^{3} \frac{x}{2\sqrt{x+1}} \,dx + 4 = 6$
$3 \int_{-1}^{3} \frac{x}{2\sqrt{x+1}} \,dx = 2$
$\int_{-1}^{3} \frac{x}{2\sqrt{x+1}} \,dx = \frac{2}{3}$