What is `int tan^(-1) (sec x + tan x) dx` equal to ?

Options

A. `(pix)/(4) + (x^(2))/(4) + c`
B. `(pix)/(2) + (x^(2))/(4) + c`
C. `(pix)/(4) + (pix^(2))/(4) + c`
D. `(pix)/(4) - (x^(2))/(4) + c`

Explanation

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

### Step 1: Simplify the Argument of the Inverse Tangent
We know that:
\[ \sec x + \tan x = \tan\left(\frac{x}{2} + \frac{\pi}{4}\right) \]
Thus, we can rewrite the integral:
\[ \tan^{-1}(\sec x + \tan x) = \tan^{-1}\left(\tan\left(\frac{x}{2} + \frac{\pi}{4}\right)\right) \]

### Step 2: Apply the Inverse Tangent Function
Using the property of the inverse tangent function, we have:
\[ \tan^{-1}(\tan(\theta)) = \theta \quad \text{(for appropriate } \theta\text{)} \]
So, we can simplify:
\[ \tan^{-1}(\sec x + \tan x) = \frac{x}{2} + \frac{\pi}{4} \]

### Step 3: Substitute Back into the Integral
Now, we substitute this back into the integral:
\[ \int \tan^{-1}(\sec x + \tan x) \, dx = \int \left(\frac{x}{2} + \frac{\pi}{4}\right) \, dx \]

### Step 4: Integrate the Expression
Now, we can integrate term by term:
\[ \int \left(\frac{x}{2} + \frac{\pi}{4}\right) \, dx = \int \frac{x}{2} \, dx + \int \frac{\pi}{4} \, dx \]
Calculating these integrals:
\[ \int \frac{x}{2} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \]
\[ \int \frac{\pi}{4} \, dx = \frac{\pi}{4} x \]

### Step 5: Combine the Results
Combining the results from the integrations, we have:
\[ \int \tan^{-1}(\sec x + \tan x) \, dx = \frac{x^2}{4} + \frac{\pi}{4} x + C \]
where \( C \) is the constant of integration.

### Final Result
Thus, the final result is:
\[ \int \tan^{-1}(\sec x + \tan x) \, dx = \frac{x^2}{4} + \frac{\pi}{4} x + C \]