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The given problem involves evaluating the improper integral:

\[
\int_0^\infty \frac{1}{n+7} \, dn
\]

### Solution:
1. Rewrite the integrand:
   \[
   \frac{1}{n+7}
   \]
   is a standard integrand whose antiderivative is:
   \[
   \ln|n+7| + C
   \]

2. Integrate:
   \[
   \int \frac{1}{n+7} \, dn = \ln(n+7) + C
   \]

3. Apply the limits of integration (from \(0\) to \(\infty\)):
   - At the upper limit (\(n \to \infty\)):
     \[
     \ln(n+7) \to \ln(\infty) = \infty
     \]
   - At the lower limit (\(n = 0\)):
     \[
     \ln(0+7) = \ln(7)
     \]

4. Subtract the results:
   \[
   \int_0^\infty \frac{1}{n+7} \, dn = \infty - \ln(7)
   \]

### Conclusion:
The integral **diverges** because the logarithmic term \(\ln(n+7)\) approaches infinity as \(n \to \infty\). Hence, the result does not converge to a finite value.

Would you like further details or a different explanation?

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Here are five related questions you might find helpful:
1. What are the conditions under which an improper integral converges?
2. How do logarithmic growth rates compare with other functions in improper integrals?
3. Can we estimate the divergence rate of \(\ln(n+7)\) in this problem?
4. What happens if the limits of integration are finite instead of improper (e.g., \(0\) to \(a\))?
5. How does the behavior of \(\int_0^\infty \frac{1}{n^2+7}\,dn\) differ from this case?

**Tip:** To handle improper integrals effectively, always analyze the behavior of the integrand at the limits of integration (especially at infinity or points of discontinuity).

Hitunglah integral tertentu \(\int_0^\infty \frac{1}{n+7} \, dn\).

Learn how to solve the improper integral \(\int_0^\infty \frac{1}{n+7} \, dn\) using logarithmic functions and convergence analysis. This detailed solution explains the behavior of the integrand and why the integral diverges to infinity. Perfect for students in Grades 10-12 studying calculus.

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