∫ (x^2 + 4x + 2) / (6x) dx
Understand the Problem
The question is asking for the integral of the rational function (x^2 + 4x + 2) / (6x) with respect to x. This entails applying techniques of integration, likely involving simplifying the expression and integrating term by term.
Answer
$$ \int \frac{x^2 + 4x + 2}{6x} \, dx = \frac{x^2}{12} + \frac{2}{3} x + \frac{1}{3} \ln |x| + C $$
Answer for screen readers
The integral is
$$ \int \frac{x^2 + 4x + 2}{6x} , dx = \frac{x^2}{12} + \frac{2}{3} x + \frac{1}{3} \ln |x| + C $$
Steps to Solve
- Simplify the integrand
First, we simplify the expression inside the integral. We can rewrite the rational function:
$$ \frac{x^2 + 4x + 2}{6x} = \frac{x^2}{6x} + \frac{4x}{6x} + \frac{2}{6x} $$
This simplifies to:
$$ \frac{x}{6} + \frac{2}{3} + \frac{1}{3x} $$
- Set up the integral
Now we rewrite the integral with the simplified expression:
$$ \int \left( \frac{x}{6} + \frac{2}{3} + \frac{1}{3x} \right) dx $$
- Integrate term by term
We integrate each term separately:
-
For the first term, $\int \frac{x}{6} , dx = \frac{1}{6} \cdot \frac{x^2}{2} = \frac{x^2}{12}$
-
For the second term, $\int \frac{2}{3} , dx = \frac{2}{3} x$
-
For the third term, $\int \frac{1}{3x} , dx = \frac{1}{3} \ln |x|$
Combining these results gives:
$$ \frac{x^2}{12} + \frac{2}{3} x + \frac{1}{3} \ln |x| + C $$
where (C) is the constant of integration.
The integral is
$$ \int \frac{x^2 + 4x + 2}{6x} , dx = \frac{x^2}{12} + \frac{2}{3} x + \frac{1}{3} \ln |x| + C $$
More Information
This integral combines polynomial and logarithmic integration techniques. Each term in the simplified integrand allows for straightforward integration, culminating in a mix of polynomial and logarithmic results.
Tips
- Forgetting to apply the linearity of integrals: Some may try to combine terms incorrectly instead of integrating them separately. Always break it into simpler integrable parts.
- Misapplying logarithm properties: When dealing with fractions that involve logarithms, ensure correct application of the absolute function.