To find the equation of the curve, we need to integrate the given derivative \(\frac{dy}{dx} = \frac{1}{(x-3)^2} + x\).

First, integrate \(\frac{1}{(x-3)^2}\):

\(\int \frac{1}{(x-3)^2} \, dx = -(x-3)^{-1} + C_1\)

Next, integrate \(x\):

\(\int x \, dx = \frac{1}{2}x^2 + C_2\)

Combine the results of the integration:

\(y = -(x-3)^{-1} + \frac{1}{2}x^2 + c\)

Use the point (2, 7) to find the constant \(c\):

\(7 = -\frac{1}{(2-3)} + \frac{1}{2}(2)^2 + c\)

\(7 = 1 + 2 + c\)

\(c = 4\)

Thus, the equation of the curve is:

\(y = -(x-3)^{-1} + \frac{1}{2}x^2 + 4\)