Question

Use partial fractions to find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) 1 /x^2 − 49 dx

Answer 1

`\displaystyle \int {(1)/(x^2 - 49)} \, dx = (1)/(14) \bigg( \ln |x - 7| - \ln |x + 7| \bigg) + C`

General Formulas and Concepts:

Algebra I

Terms/Coefficients

• Expanding/Factoring

Pre-Calculus

• Partial Fraction Decomposition

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals
• [Indefinite Integrals] integration Constant 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

Explanation:

Step 1: Define

Identify

`\displaystyle \int {(1)/(x^2 - 49)} \, dx`

Step 2: Integrate Pt. 1

1. [Integrand] Factor:
   `\displaystyle \int {(1)/(x^2 - 49)} \, dx = \int {(1)/((x - 7)(x + 7))} \, dx`
2. [Integrand] Split [Partial Fraction Decomp]:
   `\displaystyle (1)/((x - 7)(x + 7)) = (A)/(x - 7) + (B)/(x + 7)`
3. Rewrite:
   `\displaystyle 1 = A(x + 7) + B(x - 7)`
4. [Decomp] Substitute in x = 7:
   `\displaystyle 1 = A(7 + 7) + B(7 - 7)`
5. Simplify:
   `\displaystyle 1 = 14A`
6. Solve:
   `\displaystyle A = (1)/(14)`
7. [Decomp] Substitute in x = -7:
   `\displaystyle 1 = A(-7 + 7) + B(-7 - 7)`
8. Simplify:
   `\displaystyle 1 = -14B`
9. Solve:
   `\displaystyle B = (-1)/(14)`
10. [Split Integrand] Substitute in variables:
    `\displaystyle (1)/((x - 7)(x + 7)) = ((1)/(14))/(x - 7) - ((1)/(14))/(x + 7)`

Step 3: Integrate Pt. 2

1. [Integral] Rewrite [Split Integrand]:
   `\displaystyle \int {(1)/(x^2 - 49)} \, dx = \int {\bigg( ((1)/(14))/(x - 7) - ((1)/(14))/(x + 7) \bigg)} \, dx`
2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   `\displaystyle \int {(1)/(x^2 - 49)} \, dx = \int {((1)/(14))/(x - 7)} \, dx - \int {((1)/(14))/(x + 7)} \, dx`
3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {(1)/(x^2 - 49)} \, dx = (1)/(14)\int {(1)/(x - 7)} \, dx - (1)/(14)\int {(1)/(x + 7)} \, dx`
4. Factor:
   `\displaystyle \int {(1)/(x^2 - 49)} \, dx = (1)/(14) \bigg( \int {(1)/(x - 7)} \, dx - \int {(1)/(x + 7)} \, dx \bigg)`

Step 4: Integrate Pt. 3

Identify variables for u-substitution.

Integral 1

1. Set u:
   `\displaystyle u = x - 7`
2. [u] Differentiate [Basic Power Rule, Derivative Properties]:
   `\displaystyle du = dx`

Integral 2

1. Set z:
   `\displaystyle z = x + 7`
2. [z] Differentiate [Basic Power Rule, Derivative Properties]:
   `\displaystyle dz = dx`

Step 5: Integrate Pt. 4

1. [Integrals] U-Substitution:
   `\displaystyle \int {(1)/(x^2 - 49)} \, dx = (1)/(14) \bigg( \int {(1)/(u)} \, du - \int {(1)/(z)} \, dz \bigg)`
2. [Integrals] Logarithmic Integration:
   `\displaystyle \int {(1)/(x^2 - 49)} \, dx = (1)/(14) \bigg( \ln |u| - \ln |z| \bigg) + C`
3. [Variables] Back-Substitute:
   `\displaystyle \int {(1)/(x^2 - 49)} \, dx = (1)/(14) \bigg( \ln |x - 7| - \ln |x + 7| \bigg) + C`

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

Unit: Integration