∫ (x² + 4x³ + 2) / (6x) dx

Question image

Understand the Problem

The question is asking to evaluate the integral of the function given as a fraction, which involves dividing a polynomial by another polynomial and integrating with respect to x.

Answer

$$ \int \frac{x^2 + 4x^3 + 2}{6x} \, dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln|x| + C $$
Answer for screen readers

$$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln|x| + C $$

Steps to Solve

  1. Simplify the Integrand To simplify the function, divide each term in the numerator by the denominator:

$$ \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} = \frac{x}{6} + \frac{2}{3} x^2 + \frac{1}{3x} $$

  1. Rewrite the Integral Now, rewrite the integral with the simplified terms:

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

  1. Integrate Each Term Individually Now, integrate each term separately:
  • For $\frac{x}{6}$: $$ \int \frac{x}{6} , dx = \frac{1}{6} \cdot \frac{x^2}{2} = \frac{x^2}{12} $$

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

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

  1. Combine the Results Now, combine all the integrated results and add the constant of integration $C$:

$$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln|x| + C $$

$$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln|x| + C $$

More Information

This integral demonstrates how you can break down a complex fraction into simpler terms, allowing for straightforward integration. The logarithmic term appears due to the integration of the $\frac{1}{x}$ function, which frequently occurs in calculus problems.

Tips

  • Omitting the constant of integration: Always remember to add the constant $C$ after integrating.
  • Forgetting to simplify the integrand: Ensure each term is divided correctly before integrating.
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