To solve the integral \( \int \frac{1}{x \sqrt{x^4 - 1}} \, dx \), we will use a trigonometric substitution. Here are the steps to find the solution:
### Step 1: Substitution
Let \( x^2 = \sec \theta \). Then, differentiating both sides gives:
\[
2x \, dx = \sec \theta \tan \theta \, d\theta
\]
From this, we can express \( dx \) as:
\[
dx = \frac{\sec \theta \tan \theta}{2x} \, d\theta
\]
Since \( x = \sqrt{\sec \theta} \), we can substitute \( x \) back into the equation:
\[
dx = \frac{\sec \theta \tan \theta}{2\sqrt{\sec \theta}} \, d\theta = \frac{\sec^{1/2} \theta \tan \theta}{2} \, d\theta
\]
### Step 2: Rewrite the Integral
Now, we need to rewrite the integral in terms of \( \theta \):
\[
\int \frac{1}{x \sqrt{x^4 - 1}} \, dx = \int \frac{1}{\sqrt{\sec \theta} \sqrt{\sec^2 \theta - 1}} \cdot \frac{\sec^{1/2} \theta \tan \theta}{2} \, d\theta
\]
Since \( \sec^2 \theta - 1 = \tan^2 \theta \), we have:
\[
\sqrt{\sec^2 \theta - 1} = \tan \theta
\]
Thus, the integral simplifies to:
\[
\int \frac{1}{\sqrt{\sec \theta} \cdot \tan \theta} \cdot \frac{\sec^{1/2} \theta \tan \theta}{2} \, d\theta = \int \frac{1}{2} \, d\theta
\]
### Step 3: Integrate
Now we can integrate:
\[
\int \frac{1}{2} \, d\theta = \frac{1}{2} \theta + C
\]
### Step 4: Back Substitute
We need to substitute back for \( \theta \). Recall that \( x^2 = \sec \theta \), so:
\[
\theta = \sec^{-1}(x^2)
\]
Thus, the final result is:
\[
\int \frac{1}{x \sqrt{x^4 - 1}} \, dx = \frac{1}{2} \sec^{-1}(x^2) + C
\]
### Final Answer
\[
\int \frac{1}{x \sqrt{x^4 - 1}} \, dx = \frac{1}{2} \sec^{-1}(x^2) + C
\]
