A curve is such that $\frac{dy}{dx} = \frac{2}{x^2} + \frac{3}{(7 - 2x)}$ and $P(3, 1)$ is a point on the curve. Find the equation of the curve.

1. Integrate the gradient function to find the general equation of the curve

$y = \int \left( \frac{2}{x^2} + \frac{3}{7 - 2x} \right) dx$
$y = \int \left( 2x^{-2} + \frac{3}{7 - 2x} \right) dx$
$y = \frac{2x^{-1}}{-1} + \frac{3 \ln(7 - 2x)}{-2} + c$
$y = -\frac{2}{x} - \frac{3}{2} \ln(7 - 2x) + c$

2. Substitute the coordinates of the given point to find the constant of integration

Substitute $x = 3$ and $y = 1$ into the equation:
$1 = -\frac{2}{3} - \frac{3}{2} \ln(7 - 2(3)) + c$
$1 = -\frac{2}{3} - \frac{3}{2} \ln(1) + c$
Since $\ln(1) = 0$:
$1 = -\frac{2}{3} + c$
$c = 1 + \frac{2}{3} = \frac{5}{3}$

3. State the final equation of the curve

$y = -\frac{2}{x} - \frac{3}{2} \ln(7 - 2x) + \frac{5}{3}$