The integral in the image is:

$\int \left( 3x^6 - \frac{2}{x^3} \right) dx$

We can solve this by integrating each term separately.

Step 1: Integrating $3x^6$

The general rule for integrating powers of $x$ is:

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$

So, for $3x^6$, we apply the formula:

$\int 3x^6 dx = 3 \cdot \frac{x^{6+1}}{6+1} = \frac{3x^7}{7}$

Step 2: Integrating $-\frac{2}{x^3}$

We can rewrite $\frac{2}{x^3}$ as $2x^{-3}$ and then apply the power rule:

$\int 2x^{-3} dx = 2 \cdot \frac{x^{-3+1}}{-3+1} = 2 \cdot \frac{x^{-2}}{-2} = -\frac{2}{2}x^{-2} = -x^{-2}$

Step 3: Combining both results

Now, we combine the two integrals:

$\int \left( 3x^6 - \frac{2}{x^3} \right) dx = \frac{3x^7}{7} - x^{-2} + C$

Thus, the final answer is:

$\boxed{\frac{3x^7}{7} - \frac{1}{x^2} + C}$

Would you like further details on this solution or clarification?

Related questions:

1. How do you integrate functions with fractional exponents?
2. How does the power rule apply to negative exponents in integrals?
3. What is the integral of $x^{-1}$?
4. How do you handle integration when faced with more complex functions?
5. What are the applications of integrals involving polynomials in physics?

Tip:

When integrating, always check if you can simplify expressions before applying rules—this helps avoid mistakes with exponents and constants.