Evaluate the integral \(integral (3x4 - 4x3 - x2 + 1)/(x2) d(2x)\) given that \(f(1) = 2\) and find \(f(-1)\).

The given function is $f(x) = \int \frac{3x^4 - 4x^3 - x^2 + 1}{x^2} d(2x)$.
First, rewrite $d(2x)$ as $2dx$.

$$f(x) = \int \frac{3x^4 - 4x^3 - x^2 + 1}{x^2} (2dx)$$

Factor out the constant 2:

$$f(x) = 2 \int \frac{3x^4 - 4x^3 - x^2 + 1}{x^2} dx$$

Divide each term in the numerator by $x^2$:

$$f(x) = 2 \int \left( \frac{3x^4}{x^2} - \frac{4x^3}{x^2} - \frac{x^2}{x^2} + \frac{1}{x^2} \right) dx$$ $$f(x) = 2 \int \left( 3x^2 - 4x - 1 + x^{-2} \right) dx$$

Integrate each term:

$$f(x) = 2 \left( \frac{3x^{2+1}}{2+1} - \frac{4x^{1+1}}{1+1} - \frac{x^{0+1}}{0+1} + \frac{x^{-2+1}}{-2+1} \right) + C$$ $$f(x) = 2 \left( \frac{3x^3}{3} - \frac{4x^2}{2} - x - \frac{x^{-1}}{1} \right) + C$$ $$f(x) = 2 \left( x^3 - 2x^2 - x - \frac{1}{x} \right) + C$$

Distribute the 2:

$$f(x) = 2x^3 - 4x^2 - 2x - \frac{2}{x} + C$$

Substitute $x=1$ into the expression for $f(x)$:

$$f(1) = 2(1)^3 - 4(1)^2 - 2(1) - \frac{2}{1} + C$$ $$2 = 2(1) - 4(1) - 2 - 2 + C$$ $$2 = 2 - 4 - 2 - 2 + C$$ $$2 = -6 + C$$

Solve for $C$:

$$C = 2 + 6$$ $$C = 8$$

Substitute the value of $C$ back into the function:

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

Substitute $x=-1$ into the function:

$$f(-1) = 2(-1)^3 - 4(-1)^2 - 2(-1) - \frac{2}{-1} + 8$$ $$f(-1) = 2(-1) - 4(1) - (-2) - (-2) + 8$$ $$f(-1) = -2 - 4 + 2 + 2 + 8$$ $$f(-1) = -6 + 4 + 8$$ $$f(-1) = -2 + 8$$ $$f(-1) = 6$$

The final answer is $\text{6}$.