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The integral provided is:

\[
\int \frac{1}{\sqrt{3 - 2x - x^2}} \, dx
\]

To solve this integral, let's start by simplifying the expression under the square root. The quadratic expression inside the square root can be rewritten in a more convenient form by completing the square.

### Step 1: Complete the square
The expression under the square root is \(3 - 2x - x^2\). Let's rewrite it:

\[
3 - 2x - x^2 = -(x^2 + 2x - 3)
\]

Next, complete the square for the quadratic expression \(x^2 + 2x\):

\[
x^2 + 2x = \left(x + 1\right)^2 - 1
\]

So,

\[
3 - 2x - x^2 = -\left[\left(x + 1\right)^2 - 1 + 3\right] = -\left[\left(x + 1\right)^2 + 2\right]
\]

Thus, the integral becomes:

\[
\int \frac{1}{\sqrt{-\left[\left(x + 1\right)^2 + 2\right]}} \, dx
\]

This can be further simplified to:

\[
\int \frac{1}{\sqrt{-1 \cdot \left[\left(x + 1\right)^2 + 2\right]}} \, dx = \int \frac{1}{\sqrt{2 - \left(x + 1\right)^2}} \, dx
\]

### Step 2: Use trigonometric substitution
Now, let's perform a trigonometric substitution to solve this integral. Let:

\[
x + 1 = \sqrt{2} \sin\theta
\]

Then,

\[
dx = \sqrt{2} \cos\theta \, d\theta
\]

Substituting into the integral:

\[
\int \frac{1}{\sqrt{2 - (\sqrt{2} \sin\theta)^2}} \cdot \sqrt{2} \cos\theta \, d\theta
\]

The expression under the square root simplifies:

\[
\int \frac{\sqrt{2} \cos\theta}{\sqrt{2(1 - \sin^2\theta)}} \, d\theta = \int \frac{\sqrt{2} \cos\theta}{\sqrt{2} \cos\theta} \, d\theta = \int d\theta
\]

So, the integral simplifies to:

\[
\theta + C
\]

### Step 3: Convert back to \(x\)
Recall the substitution \(x + 1 = \sqrt{2} \sin\theta\), hence:

\[
\theta = \arcsin\left(\frac{x + 1}{\sqrt{2}}\right)
\]

Therefore, the integral solution is:

\[
\arcsin\left(\frac{x + 1}{\sqrt{2}}\right) + C
\]

### Final Answer:
The integral evaluates to:

\[
\int \frac{1}{\sqrt{3 - 2x - x^2}} \, dx = \arcsin\left(\frac{x + 1}{\sqrt{2}}\right) + C
\]

Would you like more details or have any other questions? Here are some related questions:

1. How do you complete the square for any quadratic expression?
2. What are some other common trigonometric substitutions used in integration?
3. How would the integral change if the quadratic expression under the square root was different?
4. What is the significance of the constant of integration \(C\) in indefinite integrals?
5. How would you solve the integral if the quadratic expression had no real roots?

**Tip:** When dealing with quadratic expressions under a square root, completing the square is often a helpful technique to simplify the integration process.

Learn how to solve the integral ∫(1/sqrt(3 - 2x - x^2)) dx step-by-step using trigonometric substitution and completing the square. This problem covers integral calculus, trigonometric substitution, and quadratic expressions. Suitable for advanced high school or college students.

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