Integrate the function (2x³ + 6x² - 12) / (x(x+3)(x² + 3x + 4)) with respect to x.

Step 1: Factor the denominator and check for irreducible quadratic factors.
The denominator is $x(x+3)(x^2+3x+4)$.
The quadratic factor $x^2+3x+4$ has a discriminant $\Delta = b^2-4ac = 3^2 - 4(1)(4) = 9 - 16 = -7$. Since the discriminant is negative, $x^2+3x+4$ is an irreducible quadratic factor.

Step 2: Set up the partial fraction decomposition.

$$\frac{2x^3 + 6x^2 - 12}{x(x+3)(x^2+3x+4)} = \frac{A}{x} + \frac{B}{x+3} + \frac{Cx+D}{x^2+3x+4}$$

Step 3: Clear the denominators by multiplying both sides by $x(x+3)(x^2+3x+4)$.

$$2x^3 + 6x^2 - 12 = A(x+3)(x^2+3x+4) + Bx(x^2+3x+4) + (Cx+D)x(x+3)$$

Step 4: Solve for the constants A, B, C, and D.
Substitute $x=0$:

$$2(0)^3 + 6(0)^2 - 12 = A(0+3)(0^2+3(0)+4) + B(0) + (C(0)+D)(0)(0+3)$$ $$-12 = A(3)(4)$$ $$-12 = 12A \implies A = -1$$

Substitute $x=-3$:

$$2(-3)^3 + 6(-3)^2 - 12 = A(0) + B(-3)((-3)^2+3(-3)+4) + (C(-3)+D)(-3)(-3+3)$$ $$2(-27) + 6(9) - 12 = B(-3)(9-9+4)$$ $$-54 + 54 - 12 = B(-3)(4)$$ $$-12 = -12B \implies B = 1$$

Now substitute $A=-1$ and $B=1$ back into the equation:

$$2x^3 + 6x^2 - 12 = -(x+3)(x^2+3x+4) + x(x^2+3x+4) + (Cx+D)x(x+3)$$

Expand the terms:

$$2x^3 + 6x^2 - 12 = -(x^3+3x^2+4x+3x^2+9x+12) + (x^3+3x^2+4x) + (Cx+D)(x^2+3x)$$ $$2x^3 + 6x^2 - 12 = -(x^3+6x^2+13x+12) + (x^3+3x^2+4x) + (Cx^3+3Cx^2+Dx^2+3Dx)$$ $$2x^3 + 6x^2 - 12 = -x^3-6x^2-13x-12 + x^3+3x^2+4x + Cx^3+3Cx^2+Dx^2+3Dx$$

Group terms by powers of $x$:

$$2x^3 + 6x^2 - 12 = (C)x^3 + (-6+3+3C+D)x^2 + (-13+4+3D)x - 12$$ $$2x^3 + 6x^2 - 12 = Cx^3 + (3C+D-3)x^2 + (3D-9)x - 12$$

Equate the coefficients of corresponding powers of $x$:
For $x^3$: $2 = C \implies C=2$
For $x^2$: $6 = 3C+D-3$. Substitute $C=2$: $6 = 3(2)+D-3 \implies 6 = 6+D-3 \implies 6 = D+3 \implies D=3$
For $x$: $0 = 3D-9$. Substitute $D=3$: $0 = 3(3)-9 \implies 0 = 9-9 \implies 0=0$ (This confirms D).
For the constant term: $-12 = -12$ (This confirms the constant term).

So, the constants are $A=-1$, $B=1$, $C=2$, $D=3$.

Step 5: Rewrite the integral using the partial fractions.

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

Step 6: Integrate each term.

$$\int \frac{-1}{x} dx = -\ln|x|$$ $$\int \frac{1}{x+3} dx = \ln|x+3|$$

For the third integral, $\int \frac{2x+3}{x^2+3x+4} dx$, notice that the numerator $2x+3$ is the derivative of the denominator $x^2+3x+4$.
Let $u = x^2+3x+4$. Then $du = (2x+3)dx$.

$$\int \frac{1}{u} du = \ln|u| = \ln|x^2+3x+4|$$

Since $x^2+3x+4$ has a negative discriminant and a positive leading coefficient, it is always positive, so we can write $\ln(x^2+3x+4)$.

Step 7: Combine the results.

$$-\ln|x| + \ln|x+3| + \ln(x^2+3x+4) + C$$

Using logarithm properties ($\ln a + \ln b - \ln c = \ln \frac{ab}{c}$):

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