Find integral: int(x+2)/(x^(3)+27)dx

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Recognize and Decompose: Recognize the integral as a rational function and attempt to decompose it into partial fractions. We have the integral: int((x+2)/(x^3+27))dx The denominator can be factored as a sum of cubes: x^3 + 27 = (x + 3)(x^2 - 3x + 9) We will attempt to decompose the integrand into partial fractions.Set up Partial Fractions: Set up the partial fraction decomposition. We express (x+2)/(x^3+27) as A/(x+3) + (Bx+C)/(x^2-3x+9), where A, B, and C are constants to be determined. (x+2)/(x^3+27) = A/(x+3) + (Bx+C)/(x^2-3x+9) Multiply both sides by the denominator (x^3+27) to clear the fractions: x+2 = A(x^2-3x+9) + (Bx+C)(x+3)Solve for Constants: Solve for the constants A, B, and C. We will now expand the right side and collect like terms to equate the coefficients on both sides of the equation. x+2 = Ax^2 - 3Ax + 9A + Bx^2 + 3Cx + Bx + 3C Combine like terms: x+2 = (A+B)x^2 + (-3A+B+3C)x + (9A+3C) Now we equate the coefficients of the corresponding powers of x on both sides of the equation: For x^2: A+B = 0 For x: -3A+B+3C = 1 For the constant term: 9A+3C = 2Solve System of Equations: Solve the system of equations for A, B, and C. From the x^2 term: A+B = 0, so B = -A. Substitute B = -A into the equation for the x term: -3A - A + 3C = 1 -4A + 3C = 1 Now we have two equations: 1) -4A + 3C = 1 2) 9A + 3C = 2 We can solve this system of equations using substitution or elimination.Correct Calculation Mistake: Solve the system of equations using elimination. Multiply the first equation by 3 and the second equation by 4 to eliminate C: 1) -12A + 9C = 3 2) 36A + 12C = 8 Now add the two equations: -12A + 9C + 36A + 12C = 3 + 8 24A + 21C = 11 Since we have only one equation with two variables, we realize there is a mistake in our calculations. We need to correct this before proceeding.

Find integral: $\int \frac{x+2}{x^{3}+27}\,dx$

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Find integral: $\int \frac{x+2}{x^{3}+27}\,dx$