Gabriella is drawing a diagram of a ramp for science class part of the diagram Is shown.

Nutka1998 [239]

Answer:

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

General Formulas and Concepts:

Algebra I

Algebra II

Calculus

Differentiation

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

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

Integration

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

Step-by-step explanation:

*Note:

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

Step 1: Define

Identify

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx$

Step 2: Integrate Pt. 1

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

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Step 4: Integrate Pt. 3

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