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Hi, how do we do this question?​

Hi, how do we do this question?​-example-1
User Yasirkula
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1 Answer

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15 votes

Answer:


\displaystyle \int {(2x)/(3x + 1)} \, dx = (-2(ln|3x + 1| - 3x))/(9) + C

General Formulas and Concepts:

Algebra I

  • Terms/Coefficients
  • Factoring

Algebra II

  • Polynomial Long Division

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
  • Integration Constant C
  • Indefinite Integrals

Integration Rule [Reverse Power Rule]:
\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + 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

Logarithmic Integration

U-Substitution

Explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

Step 1: Define

Identify


\displaystyle \int {(2x)/(3x + 1)} \, dx

Step 2: Integrate Pt. 1

  1. [Integrand] Rewrite [Polynomial Long Division (See Attachment)]:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = \int {\bigg( (2)/(3) - (2)/(3(3x + 1)) \bigg)} \, dx
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = \int {(2)/(3)} \, dx - \int {(2)/(3(3x + 1))} \, dx
  3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = (2)/(3)\int {} \, dx - (2)/(3)\int {(1)/(3x + 1)} \, dx
  4. [1st Integral] Reverse Power Rule:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = (2)/(3)x - (2)/(3)\int {(1)/(3x + 1)} \, dx

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:
    \displaystyle u = 3x + 1
  2. [u] Differentiate [Basic Power Rule]:
    \displaystyle du = 3 \ dx

Step 4: Integrate Pt. 3

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = (2)/(3)x - (2)/(9)\int {(3)/(3x + 1)} \, dx
  2. [Integral] U-Substitution:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = (2)/(3)x - (2)/(9)\int {(1)/(u)} \, du
  3. [Integral] Logarithmic Integration:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = (2)/(3)x - (2)/(9)ln|u| + C
  4. Back-Substitute:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = (2)/(3)x - (2)/(9)ln|3x + 1| + C
  5. Factor:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = -2 \bigg( (1)/(9)ln|3x + 1| - (x)/(3) \bigg) + C
  6. Rewrite:
    \displaystyle \int {(2x)/(3x + 1)} \, dx = (-2(ln|3x + 1| - 3x))/(9) + C

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

Unit: Integration

Book: College Calculus 10e

Hi, how do we do this question?​-example-1
User Jlchereau
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