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Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on Integration.
∫x^2/2x^3+1 dx.

Respuesta :

Space

Answer:

[tex]\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{\ln \big| 2x^3 + 1 \big|}{6} + C[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

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

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:

  1. f(x) = cxⁿ
  2. 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]

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx[/tex]

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

  1. Set u:                                                                                                             [tex]\displaystyle u = 2x^3 + 1[/tex]
  2. [u] Basic Power Rules [Derivative Properties]:                                            [tex]\displaystyle du = 6x^2 \ dx[/tex]

Step 3: Integrate Pt. 2

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                [tex]\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{1}{6} \int {\frac{6x^2}{2x^3 + 1}} \, dx[/tex]
  2. [Integral] U-Substitution:                                                                               [tex]\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{1}{6} \int {\frac{1}{u}} \, du[/tex]
  3. [Integral] Logarithmic Integration:                                                               [tex]\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{\ln \big| u \big|}{6} + C[/tex]
  4. [u] Back-Substitute:                                                                                       [tex]\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{\ln \big| 2x^3 + 1 \big|}{6} + C[/tex]

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

Unit: Integration

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