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The integral provided is:

\[
\int_1^2 \frac{1}{\sqrt{1 - x}} \, dx
\]

To solve this, we will approach it using substitution.

### Step 1: Substitution
Let \( u = 1 - x \). Then,
\[
du = -dx \quad \text{or} \quad dx = -du
\]

We also need to change the limits of integration. When \( x = 1 \), \( u = 0 \), and when \( x = 2 \), \( u = -1 \).

Thus, the integral becomes:

\[
-\int_0^{-1} \frac{1}{\sqrt{u}} \, du
\]

### Step 2: Simplifying the Integral
The integral of \( \frac{1}{\sqrt{u}} \) is straightforward, as:

\[
\int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u}
\]

Now, apply this to the limits:

\[
- \left[ 2\sqrt{u} \right]_0^{-1}
\]

### Step 3: Applying Limits
First, apply the upper limit \( u = -1 \), and then the lower limit \( u = 0 \):

\[
- \left( 2\sqrt{-1} - 2\sqrt{0} \right)
\]

Here, we encounter a problem: the square root of \(-1\) introduces an imaginary number \(i\), since \( \sqrt{-1} = i \).

Thus, the result involves an imaginary part:

\[
- 2i
\]

### Final Answer:
The value of the integral is \( -2i \), indicating that the integral produces a complex number.

Would you like more details or further explanations? Here are a few related questions to explore:

1. What does the introduction of imaginary numbers imply in real integrals?
2. How does substitution work when handling definite integrals?
3. What are other common techniques for solving integrals involving square roots?
4. How can improper integrals affect the final solution?
5. Why do some integrals lead to complex numbers?

**Tip:** Always check the domain of the integrand before solving, as it can reveal potential issues such as imaginary results!

Evaluate the following integral: ∫ from 1 to 2 of 1/sqrt(1-x) dx.

Learn how to solve the integral ∫ from 1 to 2 of 1/sqrt(1-x) dx using the substitution method. This problem involves key concepts in integration, including substitution and handling complex numbers. Suitable for undergraduate students.

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