The integral presented is:

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

Step 1: Simplify the integrand

Split the fraction into separate terms:

$\frac{x^6 - 3}{x^3} = \frac{x^6}{x^3} - \frac{3}{x^3} = x^3 - 3x^{-3}$

Thus, the integral becomes:

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

Step 2: Integrate term by term

Use the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

1. For $x^3$, the integral is: $\int x^3 dx = \frac{x^{3+1}}{3+1} = \frac{x^4}{4}$

2. For $-3x^{-3}$, the integral is: $\int -3x^{-3} dx = -3 \int x^{-3} dx = -3 \cdot \frac{x^{-3+1}}{-3+1} = -3 \cdot \frac{x^{-2}}{-2} = \frac{3}{2}x^{-2}$

Step 3: Combine the results

Combine the terms and add the constant of integration $C$:

$\int \left( x^3 - 3x^{-3} \right) dx = \frac{x^4}{4} + \frac{3}{2}x^{-2} + C$

Final Answer:

$\int \frac{x^6 - 3}{x^3} dx = \frac{x^4}{4} + \frac{3}{2x^2} + C$

Would you like additional clarification or a graphical explanation of this result?

Here are five related questions to expand your understanding:

1. How do you apply the power rule when the exponent is negative?
2. What happens if the integrand contains a term like $\ln(x)$?
3. How would this integral change if there were additional terms in the numerator?
4. Can you solve a similar integral if $x$ were replaced with a trigonometric function?
5. How do definite integrals differ from indefinite integrals?

Tip: Always simplify the integrand as much as possible before integrating—it can make the process significantly easier.