How do i solve?$$ \int_{}^{}\!\frac{1}{x^2\sqrt{x^2+25}}\,\mathrm{d}x $$

To solve the integral 
\[
\int \frac{1}{x^2\sqrt{x^2+25}} \, dx,
\]
we can proceed by using a trigonometric substitution.

### Step-by-Step Solution

1. **Substitute** $ x = 5 \tan \theta $:
   - Then $ dx = 5 \sec^2 \theta \, d\theta $.
   - Notice that $ x^2 = 25 \tan^2 \theta $, so $ x^2 + 25 = 25 \sec^2 \theta $ (since $1 + \tan^2 \theta = \sec^2 \theta$).

2. **Rewrite the Integral in Terms of $\theta$**:
   Substitute $ x = 5 \tan \theta $ and $ dx = 5 \sec^2 \theta \, d\theta $:
   \[
   \int \frac{1}{x^2 \sqrt{x^2 + 25}} \, dx = \int \frac{1}{25 \tan^2 \theta \cdot 5 \sec \theta} \cdot 5 \sec^2 \theta \, d\theta.
   \]

3. **Simplify the Expression**:
   \[
   = \int \frac{5 \sec^2 \theta}{25 \tan^2 \theta \cdot 5 \sec \theta} \, d\theta.
   \]
   Cancel out terms:
   \[
   = \int \frac{\sec \theta}{5 \tan^2 \theta} \, d\theta = \frac{1}{5} \int \frac{\sec \theta}{\tan^2 \theta} \, d\theta.
   \]

4. **Rewrite Using Trigonometric Identities**:
   Recall that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$. Thus,
   \[
   \frac{\sec \theta}{\tan^2 \theta} = \frac{1/\cos \theta}{\sin^2 \theta / \cos^2 \theta} = \frac{\cos \theta}{\sin^2 \theta}.
   \]
   So the integral becomes
   \[
   = \frac{1}{5} \int \frac{\cos \theta}{\sin^2 \theta} \, d\theta = \frac{1}{5} \int \csc^2 \theta \, d\theta.
   \]

5. **Integrate**:
   The integral of $\csc^2 \theta$ is $-\cot \theta$, so we have
   \[
   = -\frac{1}{5} \cot \theta + C.
   \]

6. **Substitute Back in Terms of $x$**:
   Since $ x = 5 \tan \theta $, then $\tan \theta = \frac{x}{5}$, which implies $\cot \theta = \frac{5}{x}$.
   Thus,
   \[
   -\frac{1}{5} \cot \theta = -\frac{1}{5} \cdot \frac{5}{x} = -\frac{1}{x}.
   \]

### Final Answer
\[
\int \frac{1}{x^2\sqrt{x^2+25}} \, dx = -\frac{1}{x} + C.
\]

Would you like further details on any steps?

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Here are 5 related questions to expand your understanding:

1. How would you solve a similar integral with a different trigonometric substitution?
2. What are common trigonometric identities used for simplifying integrals?
3. How can you recognize when to use trigonometric substitution in integrals?
4. What is the geometric interpretation of trigonometric substitution?
5. How does the choice of substitution affect the final integral?

**Tip:** In integrals involving $\sqrt{x^2 + a^2}$, consider a trigonometric substitution, such as $x = a \tan \theta$, to simplify the expression.

How do I solve $ \int \frac{1}{x^2\sqrt{x^2+25}} \, dx $?

Discover the step-by-step solution to the integral $ \int \frac{1}{x^2\sqrt{x^2+25}} \, dx $ using trigonometric substitution, a technique often applied in calculus to handle complex integrals involving square roots. This method includes trigonometric identities and substitutions for integrals involving $ \sqrt{x^2 + a^2} $, making it suitable for college-level calculus students.

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