`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 1: Choose \( u \) and \( dv \)

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

### Step 2: Differentiate \( u \) and Integrate \( dv \)

Now we find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = dx \]

- Integrate \( dv \):
\[ v = \int \sec^2 x \, dx = \tan x \]

### Step 3: Apply the Integration by Parts Formula

Now, we apply the integration by parts formula:
\[ \int x \sec^2 x \, dx = uv - \int v \, du \]
Substituting the values we found:
\[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \]

### Step 4: Integrate \( \tan x \)

Now we need to integrate \( \tan x \):
\[ \int \tan x \, dx = -\log |\cos x| + C = \log |\sec x| + C \]

### Step 5: Substitute Back

Now substitute back into the equation:
\[ \int x \sec^2 x \, dx = x \tan x - \left( \log |\sec x| + C \right) \]

### Final Answer

Thus, the final answer is:
\[ \int x \sec^2 x \, dx = x \tan x - \log |\sec x| + C \]