∫ (x^2 + 4x + 7) / (6x) dx
Understand the Problem
The question is asking to evaluate the integral of the rational function given by the expression in the image, where the numerator is a polynomial and the denominator is also a polynomial in terms of x.
Answer
$$ \frac{x^2}{12} + \frac{2}{3} x + \frac{7}{6} \ln |x| + C $$
Answer for screen readers
The solution to the integral is:
$$ \frac{x^2}{12} + \frac{2}{3} x + \frac{7}{6} \ln |x| + C $$
Steps to Solve
- Rewrite the Integral
We can simplify the integral by dividing the numerator by the denominator:
$$ \int \frac{x^2 + 4x + 7}{6x} , dx = \int \left( \frac{x^2}{6x} + \frac{4x}{6x} + \frac{7}{6x} \right) , dx $$
- Simplify Each Term
Now, simplify each term in the integral:
- The first term: $$ \frac{x^2}{6x} = \frac{x}{6} $$
- The second term: $$ \frac{4x}{6x} = \frac{2}{3} $$
- The third term: $$ \frac{7}{6x} = \frac{7}{6} \cdot \frac{1}{x} $$
So, the integral can be written as:
$$ \int \left( \frac{x}{6} + \frac{2}{3} + \frac{7}{6} \cdot \frac{1}{x} \right) , dx $$
- Integrate Each Term
Now, apply the integral to each term separately:
-
For $\frac{x}{6}$: $$ \int \frac{x}{6} , dx = \frac{1}{6} \cdot \frac{x^2}{2} = \frac{x^2}{12} $$
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For $\frac{2}{3}$: $$ \int \frac{2}{3} , dx = \frac{2}{3} x $$
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For $\frac{7}{6} \cdot \frac{1}{x}$: $$ \int \frac{7}{6} \cdot \frac{1}{x} , dx = \frac{7}{6} \ln |x| $$
Combining these gives:
$$ \int \left( \frac{x}{6} + \frac{2}{3} + \frac{7}{6} \cdot \frac{1}{x} \right) , dx = \frac{x^2}{12} + \frac{2}{3} x + \frac{7}{6} \ln |x| + C $$
where $C$ is the constant of integration.
The solution to the integral is:
$$ \frac{x^2}{12} + \frac{2}{3} x + \frac{7}{6} \ln |x| + C $$
More Information
This integral processes the rational function by breaking it down into simpler components. The logarithmic term arises from the integration of $\frac{1}{x}$, which is a standard result in calculus.
Tips
- Forgetting to apply the logarithmic property correctly when integrating $\frac{1}{x}$.
- Not simplifying the terms before integrating, which can lead to mistakes.
- Forgetting to include the constant of integration $C$ at the end.