∫ (x² + 4x³ + 2) / (6x) dx
Understand the Problem
The question presents an integral that needs to be evaluated. The integral involves the function (x^2 + 4x^3 + 2) divided by (6x). To solve it, we will likely simplify the expression and then find the antiderivative.
Answer
The integral evaluates to: $$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Answer for screen readers
The evaluated integral is:
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Steps to Solve
- Simplify the Integral Expression
First, divide each term in the numerator by the denominator $6x$:
$$ \int \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} , dx $$
This simplifies to:
$$ \int \left(\frac{x}{6} + \frac{2x^2}{3} + \frac{1}{3x}\right) , dx $$
- Separate the Integral
Now, separate the integral into individual parts:
$$ \int \left(\frac{x}{6} , dx + \frac{2x^2}{3} , dx + \frac{1}{3x} , dx\right) $$
- Integrate Each Part
Now, integrate each term individually:
- 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 the Results
Combine all the results from the integrals:
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
where $C$ is the constant of integration.
The evaluated integral is:
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
More Information
This integral combines polynomial functions and logarithmic integration. Understanding how to break down a more complex integral into simpler parts is crucial in calculus.
Tips
- Forgetting to separate the integral into distinct parts.
- Misapplying antiderivative rules, especially for polynomials and logarithmic functions.
- Ignoring the absolute value for the logarithmic term.
