To solve the integral of $\frac{1}{e^{7y}}$ with respect to $y$, we can rewrite the expression to make it easier to integrate:

$\int \frac{1}{e^{7y}} \, dy = \int e^{-7y} \, dy$

Now, integrating $e^{-7y}$ with respect to $y$:

$\int e^{-7y} \, dy = \frac{e^{-7y}}{-7} = -\frac{1}{7} e^{-7y} + C$

where $C$ is the constant of integration.

Final Answer:

$\int \frac{1}{e^{7y}} \, dy = -\frac{1}{7} e^{-7y} + C$

Would you like additional details or have any questions?

Here are some related questions:

1. How do you solve integrals involving exponential functions with different constants?
2. What happens when you integrate $\frac{1}{e^{ky}}$ for a general constant $k$?
3. How do you differentiate $-\frac{1}{7} e^{-7y}$ to check this result?
4. What are the rules for integrating exponential functions with negative exponents?
5. How does this integral change if the exponent is a function of $y$, such as $e^{f(y)}$?

Tip: For integrals involving exponential terms, rewrite the term so that the exponent is easier to integrate or differentiate.