To evaluate the given integral:

$\int x^5 (x^6 - 6)^4 \, dx$

Step 1: Use substitution

Let: $u = x^6 - 6$ Then: $\frac{du}{dx} = 6x^5 \quad \text{or} \quad du = 6x^5 \, dx$

Now substitute $x^5 \, dx$ with $\frac{1}{6} du$, and the integral becomes: $\int x^5 (x^6 - 6)^4 \, dx = \int u^4 \cdot \frac{1}{6} \, du$

Step 2: Simplify and integrate

$\int u^4 \cdot \frac{1}{6} \, du = \frac{1}{6} \int u^4 \, du$ The integral of $u^4$ is: $\int u^4 \, du = \frac{u^5}{5}$

So: $\frac{1}{6} \int u^4 \, du = \frac{1}{6} \cdot \frac{u^5}{5} = \frac{u^5}{30}$

Step 3: Substitute back

Recall that $u = x^6 - 6$. Substituting back: $\frac{u^5}{30} = \frac{(x^6 - 6)^5}{30}$

Final Answer:

$\int x^5 (x^6 - 6)^4 \, dx = \frac{(x^6 - 6)^5}{30} + C$

Let me know if you would like further clarifications or a deeper breakdown!

Related Questions:

1. What is the importance of substitution in solving integrals?
2. How would the solution change if the power of $x^5$ were different?
3. How can we verify the result by differentiating it?
4. What happens if the constant term in $x^6 - 6$ changes?
5. How would this integral look in definite form?

Tip:

Always check your substitution by