Respuesta :
Answer:
[tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \ln |x - 7| - \ln |x + 7| \bigg) + C[/tex]
General Formulas and Concepts:
Algebra I
Terms/Coefficients
- Expanding/Factoring
Pre-Calculus
- Partial Fraction Decomposition
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Integration
- Integrals
- [Indefinite Integrals] integration Constant C
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
U-Substitution
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx[/tex]
Step 2: Integrate Pt. 1
- [Integrand] Factor: [tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \int {\frac{1}{(x - 7)(x + 7)}} \, dx[/tex]
- [Integrand] Split [Partial Fraction Decomp]: [tex]\displaystyle \frac{1}{(x - 7)(x + 7)} = \frac{A}{x - 7} + \frac{B}{x + 7}[/tex]
- Rewrite: [tex]\displaystyle 1 = A(x + 7) + B(x - 7)[/tex]
- [Decomp] Substitute in x = 7: [tex]\displaystyle 1 = A(7 + 7) + B(7 - 7)[/tex]
- Simplify: [tex]\displaystyle 1 = 14A[/tex]
- Solve: [tex]\displaystyle A = \frac{1}{14}[/tex]
- [Decomp] Substitute in x = -7: [tex]\displaystyle 1 = A(-7 + 7) + B(-7 - 7)[/tex]
- Simplify: [tex]\displaystyle 1 = -14B[/tex]
- Solve: [tex]\displaystyle B = \frac{-1}{14}[/tex]
- [Split Integrand] Substitute in variables: [tex]\displaystyle \frac{1}{(x - 7)(x + 7)} = \frac{\frac{1}{14}}{x - 7} - \frac{\frac{1}{14}}{x + 7}[/tex]
Step 3: Integrate Pt. 2
- [Integral] Rewrite [Split Integrand]: [tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \int {\bigg( \frac{\frac{1}{14}}{x - 7} - \frac{\frac{1}{14}}{x + 7} \bigg)} \, dx[/tex]
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: [tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \int {\frac{\frac{1}{14}}{x - 7}} \, dx - \int {\frac{\frac{1}{14}}{x + 7}} \, dx[/tex]
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14}\int {\frac{1}{x - 7}} \, dx - \frac{1}{14}\int {\frac{1}{x + 7}} \, dx[/tex]
- Factor: [tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \int {\frac{1}{x - 7}} \, dx - \int {\frac{1}{x + 7}} \, dx \bigg)[/tex]
Step 4: Integrate Pt. 3
Identify variables for u-substitution.
Integral 1
- Set u: [tex]\displaystyle u = x - 7[/tex]
- [u] Differentiate [Basic Power Rule, Derivative Properties]: [tex]\displaystyle du = dx[/tex]
Integral 2
- Set z: [tex]\displaystyle z = x + 7[/tex]
- [z] Differentiate [Basic Power Rule, Derivative Properties]: [tex]\displaystyle dz = dx[/tex]
Step 5: Integrate Pt. 4
- [Integrals] U-Substitution: [tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \int {\frac{1}{u}} \, du - \int {\frac{1}{z}} \, dz \bigg)[/tex]
- [Integrals] Logarithmic Integration: [tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \ln |u| - \ln |z| \bigg) + C[/tex]
- [Variables] Back-Substitute: [tex]\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \ln |x - 7| - \ln |x + 7| \bigg) + C[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration