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Find the integral of 1\1+(2x)^2​

User Shankar BS
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Answer:


\displaystyle \int {(1)/(1+(2x)^2)} \, dx = (artan(2x))/(2) + C

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Antiderivatives - Integrals

Indefinite Integrals

  • Integral Constant C

U-Substitution

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

Arctrig Integration [arctangent]:
\displaystyle \int {(1)/(a^2 + u^2)} \, du = (1)/(a)arctan((u)/(a)) + C

Explanation:

Step 1: Define


\displaystyle \int {(1)/(1+(2x)^2)} \, dx

Step 2: Integrate Pt. 1

  1. [Integral] Rewrite:
    \displaystyle \int {(1)/(1^2+(2x)^2)} \, dx

Step 3: Identify Variables

Identify variables for u-substitution of arctrig.

  1. Set u:
    \displaystyle u = 2x
  2. Differentiate [Basic Power Rule]:
    \displaystyle (du)/(dx) = 1 \cdot 2x^(1 - 1)
  3. [Derivative] Simplify:
    \displaystyle (du)/(dx) = 2
  4. [Derivative] Rewrite:
    \displaystyle du = 2dx
  5. Set a:
    \displaystyle a = 1

Step 4: Integrate Pt. 2

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
    \displaystyle (1)/(2)\int {(2)/(1^2+(2x)^2)} \, dx
  2. [Integral] U-Substitution:
    \displaystyle (1)/(2)\int {(du)/(a +u^2)}
  3. [Integral] Arctrig Integration [arctangent]:
    \displaystyle (1)/(2)[(1)/(a)arctan((u)/(a))] + C
  4. [Integral] Back-Substitute:
    \displaystyle (1)/(2)[(1)/(1)arctan((2x)/(1))] + C
  5. [Integral] Divide:
    \displaystyle (1)/(2)[arctan(2x)] + C
  6. [Integral] Multiply:
    \displaystyle (arctan(2x))/(2) + C

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

Unit: Integration of Arctrig

Book: College Calculus 10e

User Bill Comer
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