Use partial fractions to calculate the following integrals: 1 ∫ (x+3)(x-4)/(2x³-x²-6x) dx 2 ∫ (3x²+16x+26)/((x-2)(x²+3x+4)) dx

Here are the solutions to the integrals using partial fractions. 3.1 Use partial fractions to calculate ((x+3)(x-4))/(2x^3 - x^2 - 6x) dx. Step 1: Simplify the numerator and factor the denominator. The numerator is (x+3)(x-4) = x^2 - x - 12. The denominator is 2x^3 - x^2 - 6x = x(2x^2 - x - 6). Factor the quadratic 2x^2 - x - 6: 2x^2 - x - 6 = 2x^2 - 4x + 3x - 6 = 2x(x-2) + 3(x-2) = (2x+3)(x-2). So the integrand is (x^2 - x - 12)/(x(2x+3)(x-2)). Step 2: Set up the partial fraction decomposition. (x^2 - x - 12)/(x(2x+3)(x-2)) = (A)/(x) + (B)/(2x+3) + (C)/(x-2) Multiply both sides by x(2x+3)(x-2): x^2 - x - 12 = A(2x+3)(x-2) + Bx(x-2) + Cx(2x+3) Step 3: Solve for the constants A, B, and C. Set x=0: 0^2 - 0 - 12 = A(2(0)+3)(0-2) + B(0)(0-2) + C(0)(2(0)+3) -12 = A(3)(-2) -12 = -6A A = 2. Set x=2: 2^2 - 2 - 12 = A(2(2)+3)(2-2) + B(2)(2-2) + C(2)(2(2)+3) 4 - 2 - 12 = 0 + 0 + C(2)(7) -10 = 14C C = -(10)/(14) = -(5)/(7). Set x = -(3)/(2): (-(3)/(2))^2 - (-(3)/(2)) - 12 = A() + B(-(3)/(2))(-(3)/(2)-2) + C() (9)/(4) + (3)/(2) - 12 = B(-(3)/(2))(-(7)/(2)) (9+6-48)/(4) = B((21)/(4)) -(33)/(4) = (21)/(4)B B = -(33)/(21) = -(11)/(7). Step 4: Integrate the partial fractions. ((2)/(x) - (11)/(7(2x+3)) - (5)/(7(x-2))) dx = 2 (1)/(x) dx - (11)/(7) (1)/(2x+3) dx - (5)/(7) (1)/(x-2) dx = 2 |x| - (11)/(7) · (1)/(2) |2x+3| - (5)/(7) |x-2| + C 2 |x| - (11)/(14) |2x+3| - (5)/(7) |x-2| + C 3.2 Use partial fractions to calculate (3x^2 + 16x + 26)/((x-2)(x^2 + 3x + 4)) dx. Step 1: Check the irreducibility of the quadratic factor. For the quadratic factor x^2 + 3x + 4, the discriminant is = b^2 - 4ac = 3^2 - 4(1)(4) = 9 - 16 = -7. Since the discriminant is negative, x^2 + 3x + 4 is irreducible over real numbers. Step 2: Set up the partial fraction decomposition. (3x^2 + 16x + 26)/((x-2)(x^2 + 3x + 4)) = (A)/(x-2) + (Bx+C)/(x^2+3x+4) Multiply both sides by (x-2)(x^2+3x+4): 3x^2 + 16x + 26 = A(x^2+3x+4) + (Bx+C)(x-2) Step 3: Solve for the constants A, B, and C. Set x=2: 3(2)^2 + 16(2) + 26 = A(2^2+3(2)+4) + (B(2)+C)(2-2) 12 + 32 + 26 = A(4+6+4) + 0 70 = 14A A = 5. Expand the right side and equate coefficients: 3x^2 + 16x + 26 = Ax^2+3Ax+4A + Bx^2 - 2Bx + Cx - 2C 3x^2 + 16x + 26 = (A+B)x^2 + (3A-2B+C)x + (4A-2C) Equating coefficients of x^2: A+B = 3 5+B=3 B = -2. Equating constant terms: 4A-2C = 26 4(5)-2C=26 20-2C=26 -2C=6 C = -3. (To verify, check the coefficient of x: 3A-2B+C = 3(5)-2(-2)+(-3) = 15+4-3 = 16. This matches the left side.) Step 4: Integrate the partial fractions. ((5)/(x-2) + (-2x-3)/(x^2+3x+4)) dx = 5 (1)/(x-2) dx - (2x+3)/(x^2+3x+4) dx The first integral is 5 |x-2|. For the second integral, I_2 = (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. I_2 = (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 absolute value is not strictly needed). Step 5: Combine the results. 5 |x-2| - (x^2+3x+4) + C 3 done, 2 left today. You're making progress.