Question

Use integration by parts to find the integrals in Exercise.
∫(6x+3)e-2x dx.

Answer 1

`\displaystyle \int {(6x + 3)e^(-2x)} \, dx = -3e^(-2x) (x + 1) + C`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

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`

U-Substitution

Integration by Parts:
`\displaystyle \int {u} \, dv = uv - \int {v} \, du`

• [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig

Explanation:

Step 1: Define

Identify

`\displaystyle \int {(6x + 3)e^(-2x)} \, dx`

Step 2: Integrate Pt. 1

1. [Integrand] Rewrite [Factor]:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = \int {3(2x + 1)e^(-2x)} \, dx`
2. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = 3 \int {(2x + 1)e^(-2x)} \, dx`

Step 3: Integrate Pt. 2

Identify variables for integration by parts using LIPET.

1. Set u:
   `\displaystyle u = 2x + 1`
2. [u] Basic Power Rule [Derivative Properties]:
   `\displaystyle du = 2 \ dx`
3. Set dv:
   `\displaystyle dv = e^(-2x) \ dx`
4. [dv] Exponential Integration [U-Substitution]:
   `\displaystyle v = (-e^(-2x))/(2)`

Step 4: Integrate Pt. 3

1. [Integral] Integration by Parts:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = 3 \bigg( (-(2x + 1)e^(-2x))/(2) - \int {-e^(-2x)} \, dx \bigg)`
2. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = 3 \bigg( (-(2x + 1)e^(-2x))/(2) + \int {e^(-2x)} \, dx \bigg)`

Step 5: Integrate Pt. 4

Identify variables for u-substitution.

1. Set u:
   `\displaystyle u = -2x`
2. [u] Basic Power Rule [Derivative Property - Multiplied Constant]:
   `\displaystyle du = -2 \ dx`

Step 6: Integrate Pt. 5

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = 3 \bigg( (-(2x + 1)e^(-2x))/(2) - (1)/(2) \int {-2e^(-2x)} \, dx \bigg)`
2. [Integral] U-Substitution:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = 3 \bigg( (-(2x + 1)e^(-2x))/(2) - (1)/(2) \int {e^(u)} \, du \bigg)`
3. [Integral] Exponential Integration:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = 3 \bigg( (-(2x + 1)e^(-2x))/(2) - (e^(u))/(2) \bigg) + C`
4. [u] Back-Substitute:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = 3 \bigg( (-(2x + 1)e^(-2x))/(2) - (e^(-2x))/(2) \bigg) + C`
5. Factor:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = -3e^(-2x) \bigg( (2x + 1)/(2) + (1)/(2) \bigg) + C`
6. Simplify:
   `\displaystyle \int {(6x + 3)e^(-2x)} \, dx = -3e^(-2x) (x + 1) + C`

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

Unit: Integration