`int " sec (tan"^(_1)" x ) dx "`

Answer

To solve the integral \( I = \int \sec(\tan^{-1} x) \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = \tan^{-1} x \). Then, we have:
\[ x = \tan t \quad \text{and} \quad dx = \sec^2 t \, dt \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we get:
\[ I = \int \sec(t) \sec^2(t) \, dt \]
This simplifies to:
\[ I = \int \sec^3(t) \, dt \]

### Step 3: Use the Identity
We know that:
\[ \sec^3(t) = \sec(t) \cdot \sec^2(t) \]
Using the identity \( \sec^2(t) = 1 + \tan^2(t) \), we can express:
\[ I = \int \sec(t)(1 + \tan^2(t)) \, dt \]
This can be split into two integrals:
\[ I = \int \sec(t) \, dt + \int \sec(t) \tan^2(t) \, dt \]

### Step 4: Solve Each Integral
1. The integral \( \int \sec(t) \, dt \) is known to be:
\[ \int \sec(t) \, dt = \ln | \sec(t) + \tan(t) | + C_1 \]
2. For the integral \( \int \sec(t) \tan^2(t) \, dt \), we can use the identity \( \tan^2(t) = \sec^2(t) - 1 \):
\[ \int \sec(t) \tan^2(t) \, dt = \int \sec(t)(\sec^2(t) - 1) \, dt = \int \sec^3(t) \, dt - \int \sec(t) \, dt \]

### Step 5: Combine Results
Thus, we can express \( I \) as:
\[ I = \ln | \sec(t) + \tan(t) | + C_1 + (I - \int \sec(t) \, dt) \]
This leads us to:
\[ 2I = \ln | \sec(t) + \tan(t) | + C_1 \]
So,
\[ I = \frac{1}{2} \ln | \sec(t) + \tan(t) | + \frac{C_1}{2} \]

### Step 6: Back Substitute
Now, substituting back \( t = \tan^{-1} x \):
\[ \sec(t) = \sqrt{1 + \tan^2(t)} = \sqrt{1 + x^2} \]
\[ \tan(t) = x \]
Thus, we have:
\[ I = \frac{1}{2} \ln | \sqrt{1 + x^2} + x | + C \]

### Final Result
The final answer is:
\[ I = \frac{1}{2} \ln | \sqrt{1 + x^2} + x | + C \]