Answer:
\(\displaystyle \int {\frac{x^4 - 5x^3 + 6x^2 - 18}{x^3 + 3x^2}} \, dx = 28ln|x + 3| + 2ln|x| + \frac{x^2}{2} + \frac{6}{x} - 8x + C\)
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
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]: \(\displaystyle b^{-m} = \frac{1}{b^m}\)
Algebra II
Pre-Calculus
- Partial Fraction Decomposition
Calculus
Differentiation
- Derivatives
- Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Integration
- Integrals
- Indefinite Integrals
- Integration Constant C
Integration Rule [Reverse Power Rule]: \(\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C\)
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
Step-by-step explanation:
Step 1: Define
Identify
\(\displaystyle \int {\frac{x^4 - 5x^3 + 6x^2 - 18}{x^3 + 3x^2}} \, dx\)
Step 2: Partial Fraction Decomposition
- [Integrand] Factor: \(\displaystyle \int {\frac{x^4 - 5x^3 + 6x^2 - 18}{x^2(x + 3)}} \, dx\)
- [Integrand] Simplify [Long Division, See Attachment]: \(\displaystyle \frac{x^4 - 5x^3 + 6x^2 - 18}{x^2(x + 3)} = x - 8 + \frac{30x^2 - 18}{x^2(x + 3)}\)
- Split: \(\displaystyle \frac{30x^2 - 18}{x^2(x + 3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 3}\)
- Simplify [Common Denominator]: \(\displaystyle 30x^2 - 18 = Ax(x + 3) + B(x + 3) + Cx^2\)
- [Decomp] Substitute in x = -3: \(\displaystyle 30(-3)^2 - 18 = A(-3)(-3 + 3) + B(-3 + 3) + C(-3)^2\)
- Simplify: \(\displaystyle 252 = 9C\)
- Solve: \(\displaystyle C = 28\)
- [Decomp] Substitute in x = 0: \(\displaystyle 30(0)^2 - 18 = A(0)(0 + 3) + B(0 + 3) + C(0)^2\)
- Simplify: \(\displaystyle -18 = 3B\)
- Solve: \(\displaystyle B = -6\)
- [Decomp] Coefficient Method: \(\displaystyle 3Ax + Bx = 0x\)
- [Coefficient Method] Substitute in B: \(\displaystyle 3A - 6 = 0\)
- [Coefficient Method] Solve: \(\displaystyle A = 2\)
- [Split Integrand] Substitute in variables: \(\displaystyle \frac{x^4 - 5x^3 + 6x^2 - 18}{x^2(x + 3)} = x - 8 + \frac{2}{x} - \frac{6}{x^2} + \frac{28}{x + 3}\)
Step 3: Integrate Pt. 1
- [Integral] Substitute in integrand [Split Integrand]: \(\displaystyle \int {\bigg( x - 8 + \frac{2}{x} - \frac{6}{x^2} + \frac{28}{x + 3} \bigg)} \, dx\)
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: \(\displaystyle \int {x} \, dx - \int {8} \, dx + \int {\frac{2}{x}} \, dx - \int {\frac{6}{x^2}} \, dx + \int {\frac{28}{x + 3}} \, dx\)
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: \(\displaystyle \int {x} \, dx - 8\int {} \, dx + 2\int {\frac{1}{x}} \, dx - 6\int {\frac{1}{x^2}} \, dx + 28\int {\frac{1}{x + 3}} \, dx\)
- [4th Integrand] Rewrite [Exponential Rule - Rewrite]: \(\displaystyle \int {x} \, dx - 8\int {} \, dx + 2\int {\frac{1}{x}} \, dx - 6\int {x^{-2}} \, dx + 28\int {\frac{1}{x + 3}} \, dx\)
- [1st/2nd/4th Integral] Reverse Power Rule: \(\displaystyle \frac{x^2}{2} - 8x + 2\int {\frac{1}{x}} \, dx - 6(\frac{-1}{x}) + 28\int {\frac{1}{x + 3}} \, dx\)
Step 4: Integrate Pt. 2
Identify variables for u-substitution.
- Set u: \(\displaystyle u = x + 3\)
- [u] Differentiate [Basic Power Rule]: \(\displaystyle du = dx\)
Step 5: Integrate Pt. 3
- [2nd Integral] U-Substitution: \(\displaystyle \frac{x^2}{2} - 8x + 2\int {\frac{1}{x}} \, dx - 6(\frac{-1}{x}) + 28\int {\frac{1}{u}} \, du\)
- [Integrals] Logarithmic Integration: \(\displaystyle \frac{x^2}{2} - 8x + 2ln|x| - 6(\frac{-1}{x}) + 28ln|u| + C\)
- Back-Substitute: \(\displaystyle \frac{x^2}{2} - 8x + 2ln|x| - 6(\frac{-1}{x}) + 28ln|x + 3| + C\)
- Simplify: \(\displaystyle 28ln|x + 3| + 2ln|x| + \frac{x^2}{2} + \frac{6}{x} - 8x + C\)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
