The equation of a curve is given by $y = 5xe^{2x+1}$.

1. Apply the product rule to differentiate the function

Let $u = 5x$ and $v = e^{2x+1}$.
$\frac{du}{dx} = 5$
$\frac{dv}{dx} = 2e^{2x+1}$

2. Combine using the product rule formula and simplify

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$
$\frac{dy}{dx} = (5x)(2e^{2x+1}) + (e^{2x+1})(5)$
$\frac{dy}{dx} = 10xe^{2x+1} + 5e^{2x+1}$
$\frac{dy}{dx} = 5e^{2x+1}(2x + 1)$

1. State the derivative from part (a) and express integration as the reverse process.

From (a), we have $\frac{d}{dx}(5xe^{2x+1}) = 5e^{2x+1} + 10xe^{2x+1}$.
Integrating both sides with respect to $x$:
$\int (5e^{2x+1} + 10xe^{2x+1}) \text{ d}x = 5xe^{2x+1} + c$

2. Rearrange the equation to isolate the integral of the required term.

$\int 5e^{2x+1} \text{ d}x + 10 \int xe^{2x+1} \text{ d}x = 5xe^{2x+1} + c$
$10 \int xe^{2x+1} \text{ d}x = 5xe^{2x+1} - \int 5e^{2x+1} \text{ d}x$

3. Evaluate the remaining integral and divide by 10 to find the final answer.

$10 \int xe^{2x+1} \text{ d}x = 5xe^{2x+1} - \frac{5}{2}e^{2x+1} + c$
$\int xe^{2x+1} \text{ d}x = \frac{1}{2}xe^{2x+1} - \frac{1}{4}e^{2x+1} + C$, where $C = \frac{c}{10}$

1. Set up the inequality for a decreasing function using the derivative from part (a)

From part (a), $\frac{dy}{dx} = 5e^{2x+1}(1 + 2x)$.
For $y$ to be a decreasing function, $\frac{dy}{dx} < 0$.
$5e^{2x+1}(1 + 2x) < 0$

2. Solve the inequality by considering the sign of the exponential term

Since $5e^{2x+1} > 0$ for all real values of $x$, we must have:
$1 + 2x < 0$
$2x < -1$
$x < -\frac{1}{2}$