To find the antiderivative of $\frac{1}{x^2 \sqrt{x^2 + 25}}$, we can proceed by using a trigonometric substitution to simplify the integral.

Given: $\int \frac{1}{x^2 \sqrt{x^2 + 25}} \, dx$

Let's use the substitution: $x = 5 \tan \theta$ Then, $dx = 5 \sec^2 \theta \, d\theta$, and $x^2 + 25 = 25 \sec^2 \theta.$

Substituting into the integral:

1. Rewrite $x^2 + 25$: $\sqrt{x^2 + 25} = \sqrt{25 \sec^2 \theta} = 5 \sec \theta.$

2. Substitute $x$ and $dx$: $\frac{1}{x^2 \sqrt{x^2 + 25}} = \frac{1}{25 \tan^2 \theta \cdot 5 \sec \theta} = \frac{1}{125} \cdot \frac{\cos \theta}{\sin^2 \theta}.$

Thus, the integral becomes: $\int \frac{1}{x^2 \sqrt{x^2 + 25}} \, dx = \frac{1}{125} \int \frac{\cos \theta}{\sin^2 \theta} \, d\theta.$

3. Simplify the Integral: $\int \frac{\cos \theta}{\sin^2 \theta} \, d\theta = \int \csc^2 \theta \, d\theta.$

The antiderivative of $\csc^2 \theta$ is $-\cot \theta$, so: $\int \csc^2 \theta \, d\theta = -\cot \theta.$

4. Substitute Back: Since $x = 5 \tan \theta$, we have $\cot \theta = \frac{5}{x}$. Therefore: $\int \frac{1}{x^2 \sqrt{x^2 + 25}} \, dx = -\frac{1}{125} \cdot \frac{5}{x} = -\frac{1}{25x}.$

So, the antiderivative is: $-\frac{1}{25x} + C.$

Would you like additional details or have any questions?

---

Here are five related questions to consider:

1. How would we approach this integral if the expression inside the square root were different?
2. Can we evaluate this integral using hyperbolic trigonometric substitution?
3. How does the process change if we are asked for a definite integral with bounds?
4. What happens if the power of $x$ changes from $x^2$ to a different exponent?
5. Can this type of integral be solved using a different method, such as integration by parts?

Tip: For integrals involving $\sqrt{x^2 + a^2}$, try trigonometric substitutions using $x = a \tan \theta$ for simplification.