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We are tasked with evaluating the integral:

\[
\int_1^6 \ln(t) \, dt
\]

To solve this, let's first compute the exact value of the integral. The integral of \(\ln(t)\) is solved using integration by parts. Let:

- \( u = \ln(t) \) so that \( du = \frac{1}{t} dt \),
- \( dv = dt \) so that \( v = t \).

Using the formula for integration by parts:

\[
\int u \, dv = uv - \int v \, du
\]

we get:

\[
\int \ln(t) \, dt = t \ln(t) - t + C
\]

Now, applying the limits from 1 to 6:

\[
\left[ t \ln(t) - t \right]_1^6 = \left( 6 \ln(6) - 6 \right) - \left( 1 \ln(1) - 1 \right)
\]

We know that \(\ln(1) = 0\), so the expression simplifies to:

\[
6 \ln(6) - 6 - (-1) = 6 \ln(6) - 5
\]

Thus, the exact value of the integral is:

\[
6 \ln(6) - 5
\]

### Numerical Evaluation:

We can now compute the numerical value of this expression. Using \(\ln(6) \approx 1.791759\):

\[
6 \ln(6) - 5 \approx 6(1.791759) - 5 = 10.750554 - 5 = 5.750554
\]

Rounding to three decimal places:

\[
\boxed{5.751}
\]

### Final Answers:
- **Exact answer:** \( 6 \ln(6) - 5 \)
- **Numerical answer:** \( 5.751 \)

Would you like more details on the steps or a breakdown of the integration by parts process?

Here are some related questions you might want to explore:
1. How do you apply integration by parts to different functions?
2. What are the properties of the natural logarithm in calculus?
3. Can this integral be solved using other methods, like numerical integration?
4. How does changing the bounds of integration affect the result?
5. What is the significance of the natural logarithm in applied mathematics?

**Tip:** Always verify whether the logarithmic function is properly defined on the interval when integrating, especially when there are lower bounds near zero!

Evaluate the integral ∫₁⁶ ln(t) dt both exactly and numerically.

Learn how to evaluate the integral ∫₁⁶ ln(t) dt using the integration by parts method. The exact answer is 6 ln(6) - 5, and the numerical answer is approximately 5.751. This problem is suitable for undergraduate calculus students.

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