We need to evaluate the definite integral: $\int_{-2}^{3} \frac{(x-2)(6x^2 - x - 2)}{(2x+1)} dx$.

1. Problem Description

We need to evaluate the definite integral:

$\int_{-2}^{3} \frac{(x-2)(6x^2 - x - 2)}{(2x+1)} dx$.

2. Solution Steps

First, we factor the quadratic expression $6x^2 - x - 2$. We are looking for two numbers that multiply to $6(-2) = -12$ and add up to $-1$. These numbers are $-4$ and $3$.

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

Thus, the integral becomes:

$\int_{-2}^{3} \frac{(x-2)(2x+1)(3x-2)}{(2x+1)} dx$.

We can cancel out the $(2x+1)$ term, provided that $2x+1 \neq 0$, which means $x \neq -\frac{1}{2}$. Since $-\frac{1}{2}$ is between $-2$ and $3$, we should be careful, but we proceed with the simplification and will deal with the singularity if necessary.

So, we are left with:

$\int_{-2}^{3} (x-2)(3x-2) dx$.

Expanding the expression, we have:

$\int_{-2}^{3} (3x^2 - 2x - 6x + 4) dx = \int_{-2}^{3} (3x^2 - 8x + 4) dx$.

Now, we find the antiderivative:

$\int (3x^2 - 8x + 4) dx = x^3 - 4x^2 + 4x + C$.

We evaluate the definite integral:

$[x^3 - 4x^2 + 4x]_{-2}^{3} = (3^3 - 4(3^2) + 4(3)) - ((-2)^3 - 4(-2)^2 + 4(-2)) = (27 - 36 + 12) - (-8 - 16 - 8) = (3) - (-32) = 3 + 32 = 35$.

Since the function is not defined at $x = -1/2$, the integral might not exist. Let's split the integral at this point.

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

But both integrals should lead to the same value with the evaluation process as we performed. Also, we did not get an infinite answer. Therefore, it looks like the removable singularity does not affect the value of the integral.

Therefore, we can evaluate the definite integral to be

3

5.

3. Final Answer

35