Question

Calculus - Partial Fractions

Can someone please demonstrate the indefinite integral of (x^4-5x^3+6x^2-18)/(x^3+3x^2) dx
Thank you

Answer 1

`\displaystyle \int {(x^4 - 5x^3 + 6x^2 - 18)/(x^3 + 3x^2)} \, dx = 28ln|x + 3| + 2ln|x| + (x^2)/(2) + (6)/(x) - 8x + C`

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Distributive Property

Equality Properties

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtraction Property of Equality

Algebra I

• Terms/Coefficients
• Factoring
• Exponential Rule [Rewrite]:
  `\displaystyle b^(-m) = (1)/(b^m)`

Algebra II

• Long Division

Pre-Calculus

• Partial Fraction Decomposition

Calculus

Differentiation

• Derivatives
• Derivative Notation

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals
• Indefinite Integrals
• Integration Constant C

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C`

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

Integration Property [Addition/Subtraction]:
`\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx`

U-Substitution

Logarithmic Integration

Explanation:

Step 1: Define

Identify

`\displaystyle \int {(x^4 - 5x^3 + 6x^2 - 18)/(x^3 + 3x^2)} \, dx`

Step 2: Partial Fraction Decomposition

1. [Integrand] Factor:
   `\displaystyle \int {(x^4 - 5x^3 + 6x^2 - 18)/(x^2(x + 3))} \, dx`
2. [Integrand] Simplify [Long Division, See Attachment]:
   `\displaystyle (x^4 - 5x^3 + 6x^2 - 18)/(x^2(x + 3)) = x - 8 + (30x^2 - 18)/(x^2(x + 3))`
3. Split:
   `\displaystyle (30x^2 - 18)/(x^2(x + 3)) = (A)/(x) + (B)/(x^2) + (C)/(x + 3)`
4. Simplify [Common Denominator]:
   `\displaystyle 30x^2 - 18 = Ax(x + 3) + B(x + 3) + Cx^2`
5. [Decomp] Substitute in x = -3:
   `\displaystyle 30(-3)^2 - 18 = A(-3)(-3 + 3) + B(-3 + 3) + C(-3)^2`
6. Simplify:
   `\displaystyle 252 = 9C`
7. Solve:
   `\displaystyle C = 28`
8. [Decomp] Substitute in x = 0:
   `\displaystyle 30(0)^2 - 18 = A(0)(0 + 3) + B(0 + 3) + C(0)^2`
9. Simplify:
   `\displaystyle -18 = 3B`
10. Solve:
    `\displaystyle B = -6`
11. [Decomp] Coefficient Method:
    `\displaystyle 3Ax + Bx = 0x`
12. [Coefficient Method] Substitute in B:
    `\displaystyle 3A - 6 = 0`
13. [Coefficient Method] Solve:
    `\displaystyle A = 2`
14. [Split Integrand] Substitute in variables:
    `\displaystyle (x^4 - 5x^3 + 6x^2 - 18)/(x^2(x + 3)) = x - 8 + (2)/(x) - (6)/(x^2) + (28)/(x + 3)`

Step 3: Integrate Pt. 1

1. [Integral] Substitute in integrand [Split Integrand]:
   `\displaystyle \int {\bigg( x - 8 + (2)/(x) - (6)/(x^2) + (28)/(x + 3) \bigg)} \, dx`
2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   `\displaystyle \int {x} \, dx - \int {8} \, dx + \int {(2)/(x)} \, dx - \int {(6)/(x^2)} \, dx + \int {(28)/(x + 3)} \, dx`
3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {x} \, dx - 8\int {} \, dx + 2\int {(1)/(x)} \, dx - 6\int {(1)/(x^2)} \, dx + 28\int {(1)/(x + 3)} \, dx`
4. [4th Integrand] Rewrite [Exponential Rule - Rewrite]:
   `\displaystyle \int {x} \, dx - 8\int {} \, dx + 2\int {(1)/(x)} \, dx - 6\int {x^(-2)} \, dx + 28\int {(1)/(x + 3)} \, dx`
5. [1st/2nd/4th Integral] Reverse Power Rule:
   `\displaystyle (x^2)/(2) - 8x + 2\int {(1)/(x)} \, dx - 6((-1)/(x)) + 28\int {(1)/(x + 3)} \, dx`

Step 4: Integrate Pt. 2

Identify variables for u-substitution.

1. Set u:
   `\displaystyle u = x + 3`
2. [u] Differentiate [Basic Power Rule]:
   `\displaystyle du = dx`

Step 5: Integrate Pt. 3

1. [2nd Integral] U-Substitution:
   `\displaystyle (x^2)/(2) - 8x + 2\int {(1)/(x)} \, dx - 6((-1)/(x)) + 28\int {(1)/(u)} \, du`

1. [Integrals] Logarithmic Integration:
   `\displaystyle (x^2)/(2) - 8x + 2ln|x| - 6((-1)/(x)) + 28ln|u| + C`
2. Back-Substitute:
   `\displaystyle (x^2)/(2) - 8x + 2ln|x| - 6((-1)/(x)) + 28ln|x + 3| + C`
3. Simplify:
   `\displaystyle 28ln|x + 3| + 2ln|x| + (x^2)/(2) + (6)/(x) - 8x + C`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e