Answer:
[tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C[/tex]
General Formulas and Concepts:
Algebra I
- Terms/Coefficients
- Factoring
Algebra II
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:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Integration
- Integrals
- Integration Constant C
- Indefinite Integrals
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Logarithmic Integration
U-Substitution
Step-by-step explanation:
*Note:
You could use u-solve instead of rewriting the integrand to integrate this integral.
Step 1: Define
Identify
[tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx[/tex]
Step 2: Integrate Pt. 1
- [Integrand] Rewrite [Polynomial Long Division (See Attachment)]: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx[/tex]
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx[/tex]
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx[/tex]
- [1st Integral] Reverse Power Rule: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx[/tex]
Step 3: Integrate Pt. 2
Identify variables for u-substitution.
- Set u: [tex]\displaystyle u = 3x + 1[/tex]
- [u] Differentiate [Basic Power Rule]: [tex]\displaystyle du = 3 \ dx[/tex]
Step 4: Integrate Pt. 3
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx[/tex]
- [Integral] U-Substitution: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du[/tex]
- [Integral] Logarithmic Integration: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C[/tex]
- Back-Substitute: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C[/tex]
- Factor: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3} \bigg) + C[/tex]
- Rewrite: [tex]\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
