This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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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 = 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. (2x^3 + 6x^2 - 12)/(x(x+3)(x^2+3x+4)) = (A)/(x) + (B)/(x+3) + (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 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 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 C=2 For x^2: 6 = 3C+D-3. Substitute C=2: 6 = 3(2)+D-3 6 = 6+D-3 6 = D+3 D=3 For x: 0 = 3D-9. Substitute D=3: 0 = 3(3)-9 0 = 9-9 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. ( (-1)/(x) + (1)/(x+3) + (2x+3)/(x^2+3x+4) ) dx = (-1)/(x) dx + (1)/(x+3) dx + (2x+3)/(x^2+3x+4) dx Step 6: Integrate each term. (-1)/(x) dx = -|x| (1)/(x+3) dx = |x+3| For the third integral, (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. (1)/(u) du = |u| = |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 (x^2+3x+4). Step 7: Combine the results. -|x| + |x+3| + (x^2+3x+4) + C Using logarithm properties ( a + b - c = (ab)/(c)): |((x+3)(x^2+3x+4))/(x)| + C