∫ (x² + 4x³ + 2) / (6x) dx

Understand the Problem

The question is asking to solve the integral of the expression (x^2 + 4x^3 + 2) divided by (6x) with respect to x. This involves simplifying the integrand and then integrating term by term.

Answer

$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$

Steps to Solve

Simplify the integrand We start with the integral $$ \int \frac{x^2 + 4x^3 + 2}{6x} , dx $$ We can simplify the expression 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 term by term Now we integrate each term separately:

For $\frac{x}{6}$, the integral is: $$ \int \frac{x}{6} , dx = \frac{1}{6} \cdot \frac{x^2}{2} = \frac{x^2}{12} $$
For $\frac{2}{3}x^2$, the integral is: $$ \int \frac{2}{3}x^2 , dx = \frac{2}{3} \cdot \frac{x^3}{3} = \frac{2x^3}{9} $$
For $\frac{1}{3x}$, the integral is: $$ \int \frac{1}{3x} , dx = \frac{1}{3} \ln |x| $$

Combine the integrated terms Now we combine the results of the integrations: $$ \int \left(\frac{x}{6} + \frac{2}{3} x^2 + \frac{1}{3x}\right) , dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$ where $C$ is the constant of integration.

The final answer is: $$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$

More Information

This integral demonstrates how to simplify an integrand by splitting the fraction and integrating term by term. The logarithmic term arises from integrating a reciprocal function, which is a common result in calculus.

Tips

Forgetting to simplify the integrand before integrating.
Mixing up the integration rules for different terms, especially for powers of $x$ and the natural logarithm.
Missing the constant of integration $C$ at the end.

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