The integral in the image is:

$\int \frac{dx}{(x^2 + 4)\sqrt{x^2 + 9}}$

We will solve this step by step using substitution methods.

Step-by-Step Solution:

1. Substitution: Let $x = 2 \tan(\theta)$. Therefore, $dx = 2 \sec^2(\theta) d\theta$.

   Also, $x^2 + 4 = 4(\tan^2(\theta) + 1) = 4\sec^2(\theta)$ $x^2 + 9 = 4\tan^2(\theta) + 9 = 9\sec^2(\theta)$

2. Transform the Integral: Substituting $x = 2\tan(\theta)$ and the derived expressions: $\int \frac{dx}{(x^2 + 4)\sqrt{x^2 + 9}} = \int \frac{2\sec^2(\theta) d\theta}{4\sec^2(\theta)\cdot 3\sec(\theta)}$

3. Simplify the Integral: $= \frac{1}{6} \int \frac{\sec^2(\theta)}{\sec^3(\theta)} d\theta = \frac{1}{6} \int \frac{d\theta}{\sec(\theta)} = \frac{1}{6} \int \cos(\theta) d\theta$

4. Integrate: $= \frac{1}{6} \sin(\theta) + C$

5. Back-Substitute $\theta$: Since $x = 2\tan(\theta)$, we have: $\sin(\theta) = \frac{x}{\sqrt{x^2 + 4}}$

   Thus, the final solution becomes: $\frac{1}{6} \cdot \frac{x}{\sqrt{x^2 + 4}} + C$

The final result is: $\boxed{\frac{x}{6\sqrt{x^2 + 4}} + C}$

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

5 Related Questions:

1. What is the purpose of trigonometric substitution in integrals?
2. How would this integral change if the denominator had a different expression, such as $\sqrt{x^2 + 16}$?
3. Can this integral be solved using a different substitution method, like hyperbolic functions?
4. How can we verify the result of this integral by differentiation?
5. What is the geometric interpretation of this integral?

Tip:

Always check for symmetry in integrals; it can help simplify certain steps during the solving process.