`int " x sec"^(2) " x dx "`

Answer

To solve the integral \( \int x \sec^2 x \, dx \), we will use the method of integration by parts. The formula for integration by parts is given by:

\[ \int u \, dv = uv - \int v \, du \]

### Step-by-Step Solution:

1. **Choose \( u \) and \( dv \)**:
- Let \( u = x \) (which we will differentiate)
- Let \( dv = \sec^2 x \, dx \) (which we will integrate)

2. **Differentiate \( u \) and Integrate \( dv \)**:
- Differentiate \( u \):
\[ du = dx \]
- Integrate \( dv \):
\[ v = \int \sec^2 x \, dx = \tan x \]

3. **Apply the Integration by Parts Formula**:
- Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:
\[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \]

4. **Integrate \( \tan x \)**:
- Recall that \( \int \tan x \, dx = -\ln |\cos x| + C \) (or \( \ln |\sec x| + C \)).
- Therefore, we have:
\[ \int x \sec^2 x \, dx = x \tan x - (-\ln |\cos x|) + C \]
- Simplifying gives:
\[ \int x \sec^2 x \, dx = x \tan x + \ln |\cos x| + C \]

### Final Answer:
\[ \int x \sec^2 x \, dx = x \tan x + \ln |\cos x| + C \]