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yanalaym [24]
3 years ago
11

Solve the Anti derivative.​

Mathematics
1 answer:
Alex Ar [27]3 years ago
7 0

Answer:

\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C

General Formulas and Concepts:

<u>Algebra I</u>

  • Factoring

<u>Calculus</u>

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

Integration Property [Multiplied Constant]:                                                                \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Trig Integration:                                                                                                           \displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx<u />

<u />

<u>Step 2: Integrate Pt. 1</u>

  1. [Integral] Factor fraction denominator:                                                         \displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx
  2. [Integral] Integration Property - Multiplied Constant:                                   \displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx

<u>Step 3: Identify Variables</u>

<em>Set up u-substitution for the arctan trig integration.</em>

\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx

<u>Step 4: Integrate Pt. 2</u>

  1. [Integral] Substitute u-du:                                                                               \displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du
  2. [Integral] Trig Integration:                                                                               \displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C
  3. [Integral] Simplify:                                                                                           \displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C
  4. [integral] Multiply:                                                                                           \displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C
  5. [Integral] Back-Substitute:                                                                             \displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C

Topic: AP Calculus AB

Unit: Integrals - Arctrig

Book: College Calculus 10e

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<h3>What are interests?</h3>


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Learn more on mortgage here: brainly.com/question/22846480

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