Mathematics, 26.08.2019 00:10, hinacat87

Calculus - partial fractions
can someone demonstrate the indefinite integral of (x^4-5x^3+6x^2-18)/(x^3+3x^2) dx
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$\displaystyle \int {\frac{x^4 - 5x^3 + 6x^2 - 18}{x^3 + 3x^2}} \, dx = 28ln|x + 3| + 2ln|x| + \frac{x^2}{2} + \frac{6}{x} - 8x + C$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Distributive Property

Equality Properties

Multiplication Property of EqualityDivision Property of EqualityAddition Property of EqualitySubtraction Property of Equality

Algebra I

Terms/CoefficientsFactoringExponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Algebra II

Long Division

Pre-Calculus

Partial Fraction Decomposition

Calculus

Differentiation

DerivativesDerivative Notation

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

IntegralsIndefinite IntegralsIntegration 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