∫ (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

$$ \frac{x^2}{12} + \frac{2}{3} x + \frac{7}{6} \ln |x| + C $$

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 $\frac{x}{6}$: $$ \int \frac{x}{6} , dx = \frac{1}{6} \cdot \frac{x^2}{2} = \frac{x^2}{12} $$
For $\frac{2}{3}$: $$ \int \frac{2}{3} , dx = \frac{2}{3} x $$
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.

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