`int sec^(-1)x dx`

Answer

To solve the integral \( \int \sec^{-1}(x) \, dx \), we will use the method of integration by parts. Let's go through the steps in detail.

### Step-by-Step Solution:

1. **Identify the parts for integration by parts**:
We will let:
- \( u = \sec^{-1}(x) \) (the inverse secant function)
- \( dv = dx \)

2. **Differentiate and integrate**:
We need to find \( du \) and \( v \):
- The derivative of \( u \) is:
\[ du = \frac{1}{\sqrt{x^2 - 1}} \cdot \frac{1}{x} \, dx = \frac{1}{x \sqrt{x^2 - 1}} \, dx \]
- The integral of \( dv \) is:
\[ v = x \]

3. **Apply the integration by parts formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ \int \sec^{-1}(x) \, dx = x \sec^{-1}(x) - \int x \left(\frac{1}{x \sqrt{x^2 - 1}} \, dx\right) \]

4. **Simplify the integral**:
The integral simplifies to:
\[ \int \sec^{-1}(x) \, dx = x \sec^{-1}(x) - \int \frac{1}{\sqrt{x^2 - 1}} \, dx \]

5. **Evaluate the remaining integral**:
The integral \( \int \frac{1}{\sqrt{x^2 - 1}} \, dx \) is a standard integral, which evaluates to:
\[ \int \frac{1}{\sqrt{x^2 - 1}} \, dx = \ln \left| x + \sqrt{x^2 - 1} \right| + C \]

6. **Combine the results**:
Putting everything together, we have:
\[ \int \sec^{-1}(x) \, dx = x \sec^{-1}(x) - \ln \left| x + \sqrt{x^2 - 1} \right| + C \]

### Final Answer:
\[ \int \sec^{-1}(x) \, dx = x \sec^{-1}(x) - \ln \left| x + \sqrt{x^2 - 1} \right| + C \]