∫ (x^2 + 4x^3 + 2) / (6x) dx
Understand the Problem
The question involves evaluating the integral of the expression (x^2 + 4x^3 + 2)/(6x) with respect to x. This requires simplifying the expression before finding its integral.
Answer
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Answer for screen readers
The final answer is:
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Steps to Solve
- Simplify the Expression
We start with the integral:
$$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx $$
We can simplify the fraction by dividing each term in the numerator by (6x):
$$ \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} = \frac{x}{6} + \frac{2}{3}x^2 + \frac{1}{3x} $$
Thus, we rewrite the integral:
$$ \int \left( \frac{x}{6} + \frac{2}{3}x^2 + \frac{1}{3x} \right) , dx $$
- Integrate Each Term
Now we 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
Combining all results, we have:
$$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3}\ln |x| + C $$
where (C) is the constant of integration.
The final answer is:
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
More Information
This integral involves simplifying the expression before integrating, which is a common technique in calculus. Each term is integrated independently, simplifying the process.
Tips
- Not simplifying the expression first: It's essential to break down the fraction before integrating.
- Incorrect integration of terms: Ensure that each term is integrated correctly, especially when involving constants and logarithmic functions.
