To solve the integral of $\frac{2x}{x^2 + 9}$, we'll proceed with the following steps:

$\int \frac{2x}{x^2 + 9} \, dx$

Step 1: Recognize the structure of the integrand.

The numerator $2x$ is the derivative of the denominator $x^2 + 9$. This suggests that a simple substitution can be applied.

Step 2: Apply substitution.

Let: $u = x^2 + 9$ Then: $du = 2x \, dx$

Step 3: Rewrite the integral.

Now substitute into the integral: $\int \frac{2x \, dx}{x^2 + 9} = \int \frac{du}{u}$

Step 4: Integrate.

The integral of $\frac{1}{u}$ is a standard result: $\int \frac{du}{u} = \ln|u| + C$

Step 5: Substitute back for $u$.

Now, substitute $u = x^2 + 9$ back into the result: $\ln|x^2 + 9| + C$

Since $x^2 + 9$ is always positive, we can drop the absolute value: $\ln(x^2 + 9) + C$

Final Answer:

$\int \frac{2x}{x^2 + 9} \, dx = \ln(x^2 + 9) + C$

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

1. What if the numerator was $3x$ instead of $2x$?
2. How would the integral change if the denominator were $x^2 + 16$?
3. What are the steps for solving $\int \frac{x}{x^2 + 9} \, dx$?
4. How would you solve $\int \frac{1}{x^2 + 9} \, dx$?
5. How do we apply substitution in integrals in general?

Tip: Recognizing when the numerator is a derivative of the denominator is a helpful shortcut in integration!