∫(x² + 4x³ + 2) / (6x) dx
Understand the Problem
The question is asking to compute the integral of the expression (x² + 4x³ + 2) divided by 6x with respect to x. This requires applying integration techniques to solve the integral.
Answer
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Answer for screen readers
The result of the integral is: $$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Steps to Solve
- Simplifying the Integrand
First, rewrite the integrand by dividing each term in the numerator by the denominator: $$ \frac{x^2 + 4x^3 + 2}{6x} = \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} $$ which simplifies to: $$ = \frac{x}{6} + \frac{2}{3}x^2 + \frac{1}{3x} $$
- Setting up the Integral
Now that we have simplified the integrand, the integral can be expressed as: $$ \int \left( \frac{x}{6} + \frac{2}{3}x^2 + \frac{1}{3x} \right) dx $$
- Integrating Each Term Separately
Now integrate each term one by one:
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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 $\frac{2}{3} x^2$: $$ \int \frac{2}{3} x^2 dx = \frac{2}{3} \cdot \frac{x^3}{3} = \frac{2x^3}{9} $$
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For $\frac{1}{3x}$: $$ \int \frac{1}{3x} dx = \frac{1}{3} \ln |x| $$
- Combining the Results
Now combine all the results from 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 result of the integral is: $$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
More Information
This integral combines polynomial and logarithmic functions, illustrating the power of integration techniques. The terms were integrated individually, allowing for a straightforward combination of results.
Tips
- Forgetting to add the constant of integration ( C ) at the end.
- Misapplying the integration rules, especially for rational and logarithmic functions.
- Overlooking the simplification of the integrand before integration, leading to more complicated calculations.