The gradient of a curve C is given by dy/dx = (x^2 + 3)^2/x^3, x ≠ 0 - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 1

Given that the point (3, 20) lies on the curve, we can utilize the formula for $dy/dx$ to find a specific function:

First, we integrate $dy/dx$:

$$\int \left(x^2 + 6 + 9x^{-2}\right) dx.$$

This results in:

y = \frac{x^3}{3} + 6x + \int 9x^{-2} dx + c,$$ where the integral of $9x^{-2}$ is $-9/x$. Thus, we have:

y = \frac{x^3}{3} + 6x - \frac{9}{x} + c.$$

To find the constant $c$, we substitute the point (3, 20):

$$20 = \frac{3^3}{3} + 6(3) - \frac{9}{3} + c.$$

Calculating this gives:

$$20 = 9 + 18 - 3 + c,$$

which simplifies to:

$$20 = 24 + c ightarrow c = 20 - 24 = -4.$$

Therefore, the equation for the curve C is:

y = \frac{x^3}{3} + 6x - \frac{9}{x} - 4.$$