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The integral provided is:

\[
\int \frac{1}{\sqrt{9x^2 - 36x + 37}} \, dx
\]

This type of integral can be approached using trigonometric substitution, but first, we need to simplify the expression under the square root to make the substitution more apparent. Let's complete the square for the quadratic expression \( 9x^2 - 36x + 37 \).

### Step 1: Completing the square
Start with the quadratic expression inside the square root:

\[
9x^2 - 36x + 37
\]

Factor out the 9 from the first two terms:

\[
9(x^2 - 4x) + 37
\]

Now complete the square inside the parentheses:

\[
x^2 - 4x = (x - 2)^2 - 4
\]

Thus, the expression becomes:

\[
9((x - 2)^2 - 4) + 37 = 9(x - 2)^2 - 36 + 37 = 9(x - 2)^2 + 1
\]

So, the integral becomes:

\[
\int \frac{1}{\sqrt{9(x - 2)^2 + 1}} \, dx
\]

### Step 2: Trigonometric substitution
Now we can perform a trigonometric substitution. The form \( 9(x - 2)^2 + 1 \) suggests using the substitution:

\[
x - 2 = \frac{1}{3} \tan(\theta)
\]

This gives \( dx = \frac{1}{3} \sec^2(\theta) \, d\theta \).

Substitute these into the integral:

\[
\int \frac{1}{\sqrt{9 \left( \frac{1}{9} \tan^2(\theta) \right) + 1}} \cdot \frac{1}{3} \sec^2(\theta) \, d\theta
\]

Simplify the square root:

\[
\sqrt{9 \left( \frac{1}{9} \tan^2(\theta) \right) + 1} = \sqrt{\tan^2(\theta) + 1} = \sec(\theta)
\]

Thus, the integral becomes:

\[
\int \frac{\frac{1}{3} \sec^2(\theta)}{\sec(\theta)} \, d\theta = \frac{1}{3} \int \sec(\theta) \, d\theta
\]

### Step 3: Solving the integral
The integral of \( \sec(\theta) \) is well-known:

\[
\int \sec(\theta) \, d\theta = \ln | \sec(\theta) + \tan(\theta) |
\]

Thus, we have:

\[
\frac{1}{3} \ln | \sec(\theta) + \tan(\theta) | + C
\]

### Step 4: Back-substitution
Recall that \( x - 2 = \frac{1}{3} \tan(\theta) \), so:

\[
\tan(\theta) = 3(x - 2)
\]

And from the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \), we have:

\[
\sec(\theta) = \sqrt{1 + 9(x - 2)^2}
\]

Thus, the solution is:

\[
\frac{1}{3} \ln \left| \sqrt{1 + 9(x - 2)^2} + 3(x - 2) \right| + C
\]

This is the final answer to the integral.

---

Would you like further explanation on any step?

Here are some related questions you might find helpful:
1. How do you complete the square for quadratic expressions in integrals?
2. What are the common trigonometric substitutions used in integrals?
3. How is the integral of \( \sec(\theta) \) derived?
4. Why is trigonometric substitution helpful in integrals involving square roots?
5. How do you reverse trigonometric substitutions back to the original variable?

**Tip:** When working with square root expressions, always check if completing the square simplifies the function before proceeding with substitutions.

Learn how to solve the integral ∫ 1 / √(9x^2 - 36x + 37) dx using trigonometric substitution and completing the square. This solution covers key concepts in integral calculus and is suitable for undergraduate-level students.

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