∫ (x² + 4x³ + 2) / (6x) dx
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
- Simplify the Integrand To simplify the function, divide each term in the numerator by the denominator:
$$ \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} = \frac{x}{6} + \frac{2}{3} x^2 + \frac{1}{3x} $$
- Rewrite the Integral Now, rewrite the integral with the simplified terms:
$$ \int \left( \frac{x}{6} + \frac{2}{3} x^2 + \frac{1}{3x} \right) dx $$
- 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 $\frac{1}{3x}$: $$ \int \frac{1}{3x} , dx = \frac{1}{3} \ln|x| $$
- 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.
