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The given integral is:

\[
\int \frac{x^2 + 13}{\sqrt{x^2 + 9}} \, dx
\]

To solve this, we can break it down into parts. First, we'll separate the numerator to simplify the expression:

\[
\int \frac{x^2 + 13}{\sqrt{x^2 + 9}} \, dx = \int \frac{x^2}{\sqrt{x^2 + 9}} \, dx + \int \frac{13}{\sqrt{x^2 + 9}} \, dx
\]

Now, we'll solve each part individually:

### 1. First integral:
\[
\int \frac{x^2}{\sqrt{x^2 + 9}} \, dx
\]

For this part, we can use the substitution \( x = 3 \tan \theta \), which simplifies the integral. When \( x = 3 \tan \theta \), then:

- \( dx = 3 \sec^2 \theta \, d\theta \)
- \( x^2 + 9 = 9 \sec^2 \theta \)

Substituting these into the integral:

\[
\int \frac{9 \tan^2 \theta}{3 \sec \theta} 3 \sec^2 \theta \, d\theta = 9 \int \frac{\tan^2 \theta}{\sec \theta} \sec^2 \theta \, d\theta
\]

This simplifies to:

\[
9 \int \tan^2 \theta \, d\theta
\]

Using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \), the integral becomes:

\[
9 \int (\sec^2 \theta - 1) \, d\theta = 9 (\tan \theta - \theta)
\]

Since \( \tan \theta = \frac{x}{3} \) and \( \theta = \tan^{-1} \left( \frac{x}{3} \right) \), the first part of the integral becomes:

\[
9 \left( \frac{x}{3} - \tan^{-1} \left( \frac{x}{3} \right) \right) = 3x - 9 \tan^{-1} \left( \frac{x}{3} \right)
\]

### 2. Second integral:
\[
\int \frac{13}{\sqrt{x^2 + 9}} \, dx
\]

This is a standard integral, and the result is:

\[
13 \sinh^{-1} \left( \frac{x}{3} \right) \quad \text{or} \quad 13 \ln \left( x + \sqrt{x^2 + 9} \right)
\]

### Final Answer:
Now, combining both parts, the solution to the integral is:

\[
3x - 9 \tan^{-1} \left( \frac{x}{3} \right) + 13 \ln \left( x + \sqrt{x^2 + 9} \right) + C
\]

where \( C \) is the constant of integration.

Let me know if you'd like further details on any of the steps!

---

### Related Questions:
1. How does the substitution \( x = 3 \tan \theta \) simplify integrals involving \( \sqrt{x^2 + 9} \)?
2. What are the common integrals involving inverse hyperbolic functions?
3. Can you show how to derive the identity \( \tan^2 \theta = \sec^2 \theta - 1 \)?
4. How does one solve integrals involving \( \sqrt{a^2 + x^2} \)?
5. What are the differences between the hyperbolic and inverse trigonometric forms in integration?

### Tip:
Always try substitution when you encounter integrals involving square roots of quadratic expressions. It simplifies the expression and leads to standard forms.

Solve the integral: ∫ (x^2 + 13) / sqrt(x^2 + 9) dx.

Discover the steps to solve the integral of (x^2 + 13) / sqrt(x^2 + 9) using trigonometric substitution. This guide covers essential integral calculus concepts and trigonometric identities, suitable for university level mathematics.

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