∫ (x² + 4x³ + 2) / (6x) dx

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Understand the Problem

The question presents an integral that needs to be evaluated. The integral involves the function (x^2 + 4x^3 + 2) divided by (6x). To solve it, we will likely simplify the expression and then find the antiderivative.

Answer

The integral evaluates to: $$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Answer for screen readers

The evaluated integral is:

$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$

Steps to Solve

  1. Simplify the Integral Expression

First, divide each term in the numerator by the denominator $6x$:

$$ \int \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} , dx $$

This simplifies to:

$$ \int \left(\frac{x}{6} + \frac{2x^2}{3} + \frac{1}{3x}\right) , dx $$

  1. Separate the Integral

Now, separate the integral into individual parts:

$$ \int \left(\frac{x}{6} , dx + \frac{2x^2}{3} , dx + \frac{1}{3x} , dx\right) $$

  1. Integrate Each Part

Now, integrate each term individually:

  • For $\frac{x}{6}$:

$$ \int \frac{x}{6} , dx = \frac{1}{6} \cdot \frac{x^2}{2} = \frac{x^2}{12} $$

  • For $\frac{2x^2}{3}$:

$$ \int \frac{2x^2}{3} , dx = \frac{2}{3} \cdot \frac{x^3}{3} = \frac{2x^3}{9} $$

  • For $\frac{1}{3x}$:

$$ \int \frac{1}{3x} , dx = \frac{1}{3} \ln |x| $$

  1. Combine the Results

Combine all the results from the integrals:

$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$

where $C$ is the constant of integration.

The evaluated integral is:

$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$

More Information

This integral combines polynomial functions and logarithmic integration. Understanding how to break down a more complex integral into simpler parts is crucial in calculus.

Tips

  • Forgetting to separate the integral into distinct parts.
  • Misapplying antiderivative rules, especially for polynomials and logarithmic functions.
  • Ignoring the absolute value for the logarithmic term.
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