To solve the integral of \( 3x^{-6} - 2x^{-2} + 2x \, dx \), we will integrate each term separately using the power rule for integration:

\[
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C
\]
where \( C \) is the constant of integration, and \( n \neq -1 \).

Now, let's apply this to each term:

1. \( \int 3x^{-6} \, dx = 3 \cdot \frac{x^{-6+1}}{-6+1} = 3 \cdot \frac{x^{-5}}{-5} = -\frac{3}{5}x^{-5} \)
2. \( \int -2x^{-2} \, dx = -2 \cdot \frac{x^{-2+1}}{-2+1} = -2 \cdot \frac{x^{-1}}{-1} = 2x^{-1} \)
3. \( \int 2x \, dx = 2 \cdot \frac{x^{1+1}}{1+1} = 2 \cdot \frac{x^2}{2} = x^2 \)

Now, summing these results together:

\[
\int (3x^{-6} - 2x^{-2} + 2x) \, dx = -\frac{3}{5}x^{-5} + 2x^{-1} + x^2 + C
\]

where \( C \) is the constant of integration.

### Final Answer:
\[
-\frac{3}{5}x^{-5} + 2x^{-1} + x^2 + C
\]

Would you like more details, or do you have any questions?

Here are 5 related questions you might find useful:
1. How do you integrate functions with negative exponents in general?
2. What happens if the exponent is \( -1 \)? How do you integrate that?
3. What is the difference between power rule and other integration methods?
4. Can you apply this method to trigonometric functions like \( \sin(x) \)?
5. How do definite integrals differ from indefinite integrals?

**Tip:** Always check for any simplifications after integrating to make sure your result is in its simplest form.

Discover how to integrate the function 3x^-6 - 2x^-2 + 2x using the power rule for integration. This detailed solution breaks down the process, highlighting key calculus concepts such as negative exponents and the power rule. Perfect for high school students in Grades 11-12.

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