∫ (x^2 + 4x^3 + 2) / (6x) dx
Understand the Problem
The question is asking to solve the integral of the expression (x^2 + 4x^3 + 2) / (6x) with respect to x. This involves simplifying the expression and then applying integral calculus techniques.
Answer
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Answer for screen readers
The result of the integral is: $$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Steps to Solve
- Simplify the expression
Start by simplifying the integrand (\frac{x^2 + 4x^3 + 2}{6x}).
This can be broken down as follows: $$ \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} = \frac{x}{6} + \frac{2x^2}{3} + \frac{1}{3x} $$
- Set up the integral
Now we can express the integral with the simplified form: $$ \int \left(\frac{x}{6} + \frac{2x^2}{3} + \frac{1}{3x}\right) dx $$
- Integrate each term separately
Now we integrate each term:
- 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}): $$ \int \frac{1}{3x} , dx = \frac{1}{3} \ln |x| $$
Combine these results for the complete integral.
- Combine the results
Putting it all together: $$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
The result of the integral is: $$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
More Information
This integral combines polynomial and logarithmic functions. Each term integrates according to standard rules of integration, employing the power rule and the natural logarithm for the term involving (x).
Tips
Common mistakes include:
- Forgetting to add the constant (C) of integration at the end.
- Miscalculating individual integrals, especially with polynomial coefficients.
- Not simplifying the expression before integrating.
