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The integral you have provided is:

\[
\int \frac{dx}{9x^2 + 16x + 5}
\]

To solve this integral, we can use the method of completing the square in the denominator. Here's the step-by-step solution:

### Step 1: Complete the square in the denominator
The expression in the denominator is \(9x^2 + 16x + 5\). We can rewrite this by completing the square:

\[
9x^2 + 16x + 5 = 9\left(x^2 + \frac{16}{9}x\right) + 5
\]

Now, take half the coefficient of \(x\) inside the parentheses, square it, and add/subtract it inside the expression:

\[
9\left(x^2 + \frac{16}{9}x + \left(\frac{8}{9}\right)^2 - \left(\frac{8}{9}\right)^2\right) + 5 = 9\left(\left(x + \frac{8}{9}\right)^2 - \left(\frac{8}{9}\right)^2\right) + 5
\]

Simplifying further:

\[
9\left(x + \frac{8}{9}\right)^2 - 9\left(\frac{8}{9}\right)^2 + 5 = 9\left(x + \frac{8}{9}\right)^2 - \frac{64}{9} + 5
\]

\[
= 9\left(x + \frac{8}{9}\right)^2 - \frac{64}{9} + \frac{45}{9} = 9\left(x + \frac{8}{9}\right)^2 - \frac{19}{9}
\]

So, the denominator becomes:

\[
9\left[\left(x + \frac{8}{9}\right)^2 - \frac{19}{81}\right]
\]

### Step 2: Rewrite the integral
The integral now becomes:

\[
\int \frac{dx}{9\left[\left(x + \frac{8}{9}\right)^2 - \left(\frac{\sqrt{19}}{9}\right)^2\right]}
\]

### Step 3: Use a trigonometric substitution
Let:

\[
u = x + \frac{8}{9}
\]

This implies:

\[
du = dx
\]

The integral is now:

\[
\frac{1}{9}\int \frac{du}{\left(u^2 - \left(\frac{\sqrt{19}}{9}\right)^2\right)}
\]

This integral is of the form:

\[
\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \ln \left|\frac{u - a}{u + a}\right| + C
\]

where \( a = \frac{\sqrt{19}}{9} \). Substituting this into our integral:

\[
\frac{1}{9} \cdot \frac{1}{2 \cdot \frac{\sqrt{19}}{9}} \ln \left|\frac{u - \frac{\sqrt{19}}{9}}{u + \frac{\sqrt{19}}{9}}\right| + C
\]

Simplifying:

\[
\frac{1}{2\sqrt{19}} \ln \left|\frac{u - \frac{\sqrt{19}}{9}}{u + \frac{\sqrt{19}}{9}}\right| + C
\]

### Step 4: Substitute back the original variable
Finally, substituting \( u = x + \frac{8}{9} \) back:

\[
\frac{1}{2\sqrt{19}} \ln \left|\frac{x + \frac{8}{9} - \frac{\sqrt{19}}{9}}{x + \frac{8}{9} + \frac{\sqrt{19}}{9}}\right| + C
\]

This is the solution to the integral.

Would you like more details on any specific step or have any further questions?

Here are five related questions to expand your understanding:

1. How do you apply the method of partial fractions to evaluate integrals?
2. What is the significance of completing the square when solving integrals?
3. Can you explain the process of trigonometric substitution in integrals?
4. How do you identify the appropriate substitution to simplify an integral?
5. How would the integral change if the coefficients in the quadratic were different?

**Tip:** Always check if the quadratic in the denominator can be factored or if completing the square simplifies the integral.

Learn how to solve the integral of 1/(9x^2 + 16x + 5) using the method of completing the square. This problem covers advanced techniques in calculus, including completing the square, trigonometric substitution, and integration formulas. Suitable for college-level students and those preparing for calculus exams.

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