5. Trig Substitution. Use a trig substitution to evaluate the indefinite integral. Show all the details of
your substitution.
fdx/x^2 sqrtx^2+1
ANTIDERIVATIVE:
answers only

The integral given is:

\[
\int \frac{dx}{x^2 \sqrt{x^2 + 1}}
\]

### 1. Substitution:  
Use the substitution \( x = \tan(\theta) \), which gives:
\[
dx = \sec^2(\theta) d\theta, \quad x^2 + 1 = \sec^2(\theta)
\]

### 2. Transform the integral:
Substitute \( x = \tan(\theta) \) into the integral:
\[
\int \frac{\sec^2(\theta) d\theta}{\tan^2(\theta) \sec(\theta)}
= \int \frac{d\theta}{\tan^2(\theta)}
= \int \cot^2(\theta) d\theta
\]

### 3. Simplify and integrate:
Use the identity \( \cot^2(\theta) = \csc^2(\theta) - 1 \):
\[
\int \cot^2(\theta) d\theta = \int (\csc^2(\theta) - 1) d\theta
\]
\[
= -\cot(\theta) - \theta + C
\]

### 4. Back-substitute \( \theta = \tan^{-1}(x) \):
Since \( \cot(\theta) = \frac{1}{x} \) and \( \theta = \tan^{-1}(x) \), we have:
\[
-\frac{1}{x} - \tan^{-1}(x) + C
\]

### Final Answer:
\[
\boxed{-\frac{1}{x} - \tan^{-1}(x) + C}
\]

Would you like to review any specific details or steps?

Here are 5 related questions to explore further:
1. How does the identity \( \cot^2(\theta) = \csc^2(\theta) - 1 \) arise?
2. Can you use different substitutions for the same integral? What happens?
3. How does this method apply to similar integrals involving \( \sqrt{x^2 - 1} \)?
4. What are the conditions for using trigonometric substitution?
5. Can you derive the formulas for the inverse trigonometric functions used here?

**Tip:** Always double-check trigonometric identities when simplifying integrals involving substitutions.

5. Trig Substitution. Use a trig substitution to evaluate the indefinite integral. Show all the details of your substitution. fdx/x^2 sqrt(x^2+1)

This page covers the step-by-step solution for evaluating the integral ∫dx / x^2√(x^2 + 1) using trigonometric substitution. The problem involves applying the substitution x = tan(θ) and utilizing key trigonometric identities. Suitable for students in undergraduate calculus courses.

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