To evaluate the integral

$\int_{2}^{5} \frac{1}{\sqrt{6t + 7}} \, dt$

we can proceed with a substitution to simplify the integrand.

Step-by-Step Solution

1. Substitution: Let $u = 6t + 7$.

   Then, $\frac{du}{dt} = 6$, so $dt = \frac{du}{6}$.

2. Change of Limits:

   • When $t = 2$: $u = 6(2) + 7 = 12 + 7 = 19$.
   • When $t = 5$: $u = 6(5) + 7 = 30 + 7 = 37$.

   Thus, the integral becomes:

   $\int_{19}^{37} \frac{1}{\sqrt{u}} \cdot \frac{1}{6} \, du = \frac{1}{6} \int_{19}^{37} u^{-\frac{1}{2}} \, du$

3. Integrate:

   • The integral of $u^{-\frac{1}{2}}$ is $2u^{\frac{1}{2}}$.

   So,

   $\frac{1}{6} \int_{19}^{37} u^{-\frac{1}{2}} \, du = \frac{1}{6} \left[ 2u^{\frac{1}{2}} \right]_{19}^{37}$

   Simplify further:

   $= \frac{1}{6} \cdot 2 \left[ \sqrt{37} - \sqrt{19} \right] = \frac{1}{3} \left( \sqrt{37} - \sqrt{19} \right)$

4. Final Answer:

   $\int_{2}^{5} \frac{1}{\sqrt{6t + 7}} \, dt = \frac{1}{3} \left( \sqrt{37} - \sqrt{19} \right)$

This is the exact answer. If you need a decimal approximation, you can calculate:

$\approx \frac{1}{3} (6.08 - 4.36) \approx \frac{1}{3} \times 1.72 \approx 0.573$

Would you like more details on this process or have any questions?

Here are 5 additional questions that relate to this topic:

1. How do substitutions help simplify definite integrals?
2. What other techniques are useful for evaluating integrals with square root terms?
3. Why is it necessary to adjust the limits of integration after substitution?
4. How would this problem change if there were additional terms in the denominator?
5. Can the result of this integral be verified by any numerical methods?

Tip: Always verify your answer by back-substituting and checking if the result makes sense, especially when handling integrals with variable limits after substitution.