To solve the integral

$\int (x^3 - 6x^2 + 3) \, dx$

we’ll integrate each term individually. Let's go through the solution:

1. Integrate $x^3$: $\int x^3 \, dx = \frac{x^{4}}{4}$

2. Integrate $-6x^2$: $\int -6x^2 \, dx = -6 \cdot \frac{x^3}{3} = -2x^3$

3. Integrate $3$: $\int 3 \, dx = 3x$

Putting it all together, the integral becomes:

$\frac{x^4}{4} - 2x^3 + 3x + C$

where $C$ is the constant of integration.

So, the correct answer is:

$\frac{x^4}{4} - 2x^3 + 3x + C$

Would you like to see how to verify this by differentiation?

Additional Practice Questions

1. How would you integrate $\int (2x^3 - 4x + 5) \, dx$?
2. What is the antiderivative of $x^5$?
3. How do you verify an integral solution by differentiation?
4. If you integrate $-5x^4 + 2x$, what is the result?
5. What happens if there’s a coefficient before the integral, such as $2 \int (x^3 - 6x^2 + 3) \, dx$?

Tip:

When integrating, remember to add $+C$ at the end to account for the constant of integration.