The given integral is:

$\int_1^7 \frac{1}{\sqrt{x-1}} \, dx$

This can be solved by using a substitution. Let's go through the steps:

Step 1: Substitution

Let: $u = x - 1 \implies du = dx$

When $x = 1$, $u = 0$, and when $x = 7$, $u = 6$.

The integral transforms into: $\int_0^6 \frac{1}{\sqrt{u}} \, du$

Step 2: Solving the Integral

The integral of $\frac{1}{\sqrt{u}}$ is: $\int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u}$

Thus, we need to evaluate: $2\sqrt{u} \Bigg|_0^6$

Step 3: Evaluate at the Limits

$2\sqrt{6} - 2\sqrt{0} = 2\sqrt{6} = 2 \times 2.44949 = 4.89898$

Thus, the value of the integral is approximately:

$4.89898$

Rounded to 6 decimal places, the result is:

$4.898979$

Would you like any more details on the steps?

Here are some related questions:

1. How would you solve this if the limits were different?
2. What substitution method could be applied to other integrals?
3. What if the function inside the integral was more complex?
4. How do definite and indefinite integrals differ in interpretation?
5. What are the rules for simplifying expressions under square roots?

Tip: Substitution is a useful technique for solving integrals that involve square roots or polynomial expressions.