To solve the integral

$\int 5x \left(7x^2 - 3\right)^6 \, dx,$

we can use the substitution method. Let's set

$u = 7x^2 - 3.$

Now, we need to find $du$:

$du = \frac{d}{dx}(7x^2 - 3) \, dx = 14x \, dx.$

We can solve for $dx$:

$dx = \frac{du}{14x}.$

Now we can rewrite the integral in terms of $u$:

1. Substitute $u$ for $7x^2 - 3$.
2. Since we have $5x \, dx$, we can express it using $du$:

$5x \, dx = \frac{5}{14} du.$

Thus, our integral becomes:

$\int 5x \left(7x^2 - 3\right)^6 \, dx = \int \frac{5}{14} u^6 \, du.$

Now, we can integrate:

$\int u^6 \, du = \frac{u^7}{7} + C.$

So,

$\int \frac{5}{14} u^6 \, du = \frac{5}{14} \cdot \frac{u^7}{7} + C = \frac{5}{98} u^7 + C.$

Finally, we substitute back $u = 7x^2 - 3$:

$\frac{5}{98} (7x^2 - 3)^7 + C.$

Thus, the solution to the integral is:

$\int 5x \left(7x^2 - 3\right)^6 \, dx = \frac{5}{98} (7x^2 - 3)^7 + C.$

If you have any questions or need further details, let me know! Here are some related questions you might consider:

1. What other techniques can be used for integrals involving polynomials?
2. Can you explain the process of finding $du$ in more detail?
3. What if the integral had a different polynomial in the parentheses?
4. How would you apply integration by parts to a similar problem?
5. What are some real-world applications of integral calculus?

Tip: Always check if a substitution can simplify the integral before proceeding!