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The image shows the following integral to solve:

\[
\int \frac{8x}{\sqrt{9x^2 + 7}} \, dx
\]

### Solution:
We can use a substitution method to solve this integral. Let’s go step-by-step:

1. **Substitute:**
   Let \( u = 9x^2 + 7 \), so that \( du/dx = 18x \), or \( du = 18x \, dx \).

2. **Rewrite the integral:**
   The integral becomes:
   \[
   \int \frac{8x}{\sqrt{9x^2 + 7}} \, dx = \int \frac{8x}{\sqrt{u}} \, dx
   \]

   Now, replace \( dx \) using the substitution \( du = 18x \, dx \):
   \[
   \frac{du}{18x} = dx
   \]

   Substituting this into the equation:
   \[
   \int \frac{8x}{\sqrt{u}} \cdot \frac{du}{18x} = \frac{8}{18} \int \frac{du}{\sqrt{u}} = \frac{4}{9} \int u^{-\frac{1}{2}} \, du
   \]

3. **Integrate:**
   The integral of \( u^{-\frac{1}{2}} \) is \( 2u^{\frac{1}{2}} \). So:
   \[
   \frac{4}{9} \cdot 2u^{\frac{1}{2}} = \frac{8}{9} \sqrt{u}
   \]

4. **Substitute back:**
   Now, substitute \( u = 9x^2 + 7 \) back into the expression:
   \[
   \frac{8}{9} \sqrt{9x^2 + 7} + C
   \]

### Final Answer:
\[
\int \frac{8x}{\sqrt{9x^2 + 7}} \, dx = \frac{8}{9} \sqrt{9x^2 + 7} + C
\]

Would you like further details or have any questions?

Here are 5 related questions for deeper understanding:
1. What is the general method for solving integrals using substitution?
2. How can you determine when to use substitution versus integration by parts?
3. What other integrals can be simplified using a similar substitution method?
4. Can you perform a similar integral if the constant 7 inside the square root was different?
5. What happens to the solution if the term inside the square root was more complex (e.g., \( 9x^4 + 7 \))?

**Tip:** When solving integrals involving square roots, always check if substitution or trigonometric identities can simplify the expression.

Solve the integral: ∫ (8x / sqrt(9x^2 + 7)) dx

Learn how to solve the integral ∫ (8x / sqrt(9x^2 + 7)) dx using the substitution method. This step-by-step solution explains how to approach complex integrals with square roots in the denominator, suitable for advanced high school or university students.

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