∫ (x^2 + 4x^3 + 2) / (6x) dx

Question image

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

  1. 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:

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

  1. 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}):

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

  1. 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.
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