∫ (x² + 4x³ + 2) / (6x) dx
Understand the Problem
The question is asking to solve the integral of the expression (x² + 4x³ + 2) / (6x) with respect to x. This involves applying integration techniques to simplify and find the antiderivative.
Answer
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Answer for screen readers
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Steps to Solve
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Simplify the expression inside the integral
We start with the given integral: $$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx $$
We can simplify this by dividing each term in the numerator by (6x): $$ = \int \left( \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} \right) , dx $$
This simplifies to:
$$ = \int \left( \frac{x}{6} + \frac{2}{3}x^2 + \frac{1}{3x} \right) , dx $$ -
Integrate each term
Now we integrate term by term:
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For ( \frac{x}{6} ), the integral is: $$ \int \frac{x}{6} , dx = \frac{1}{6} \cdot \frac{x^2}{2} = \frac{x^2}{12} $$
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For ( \frac{2}{3}x^2 ), the integral is: $$ \int \frac{2}{3}x^2 , dx = \frac{2}{3} \cdot \frac{x^3}{3} = \frac{2x^3}{9} $$
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For ( \frac{1}{3x} ), the integral is: $$ \int \frac{1}{3x} , dx = \frac{1}{3} \ln |x| $$
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Combine the results
Now, we can combine the results of each integral: $$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
where (C) is the constant of integration.
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
More Information
This integral represents the antiderivative of the function ( \frac{x^2 + 4x^3 + 2}{6x} ). Each term was integrated separately, making the overall process straightforward.
Tips
- Confusing integration with differentiation: Ensure you're applying the correct rules for integration.
- Forgetting the constant of integration: Always remember to add (C) at the end.
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