Simplify the integrand by factoring the numerator.

The integrand is $\frac{(x-2)(6x^2 - x - 2)}{(2x+1)}$.
First, factor the quadratic term $6x^2 - x - 2$. We look for two numbers that multiply to $6 \times (-2) = -12$ and add to $-1$. These numbers are $-4$ and 3.

$$6x^2 - x - 2 = 6x^2 - 4x + 3x - 2$$ $$= 2x(3x - 2) + 1(3x - 2)$$ $$= (2x+1)(3x-2)$$

Now substitute this back into the integrand:

$$\frac{(x-2)(6x^2 - x - 2)}{(2x+1)} = \frac{(x-2)(2x+1)(3x-2)}{(2x+1)}$$

Assuming $2x+1 \neq 0$, we can cancel the $(2x+1)$ terms:

$$(x-2)(3x-2)$$

Expand this product:

$$(x-2)(3x-2) = x(3x-2) - 2(3x-2)$$ $$= 3x^2 - 2x - 6x + 4$$ $$= 3x^2 - 8x + 4$$

So the integral becomes:

$$\int_{-2}^{3} (3x^2 - 8x + 4) dx$$

The antiderivative of $3x^2 - 8x + 4$ is:

$$F(x) = \frac{3x^{2+1}}{2+1} - \frac{8x^{1+1}}{1+1} + 4x$$ $$F(x) = \frac{3x^3}{3} - \frac{8x^2}{2} + 4x$$ $$F(x) = x^3 - 4x^2 + 4x$$

$$\int_{-2}^{3} (3x^2 - 8x + 4) dx = [x^3 - 4x^2 + 4x]_{-2}^{3}$$ $$= (3^3 - 4(3)^2 + 4(3)) - ((-2)^3 - 4(-2)^2 + 4(-2))$$

Calculate the value at the upper limit ($x=3$):

$$3^3 - 4(3)^2 + 4(3) = 27 - 4(9) + 12$$ $$= 27 - 36 + 12$$ $$= -9 + 12 = 3$$

Calculate the value at the lower limit ($x=-2$):

$$(-2)^3 - 4(-2)^2 + 4(-2) = -8 - 4(4) - 8$$ $$= -8 - 16 - 8$$ $$= -32$$

Subtract the lower limit value from the upper limit value:

$$3 - (-32) = 3 + 32 = 35$$

The final answer is $\mathbf{35}$.