`inttan^(-1)(secx+tanx)dx=?`

Options

A. `(pix)/(4)+(x^(2))/(4)+C`
B. `(pix)/(4)-(x^(2))/(4)+C`
C. `(1)/((1+x^(2)))+C`
D. none of these

Explanation

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

### Step 1: Simplify the expression inside the integral
We start with the expression \( \tan^{-1}(\sec x + \tan x) \). We know that:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Thus, we can rewrite \( \sec x + \tan x \) as:
\[ \sec x + \tan x = \frac{1 + \sin x}{\cos x} \]
So, we have:
\[ \tan^{-1}(\sec x + \tan x) = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right) \]

### Step 2: Use a trigonometric identity
We can use the identity for \( \tan^{-1} \) to simplify further. We know:
\[ \tan^{-1}(a) = \tan^{-1}\left(\frac{b}{c}\right) \quad \text{where } a = \frac{b}{c} \]
This gives us:
\[ \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right) \]
This is because:
\[ \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4} \tan\frac{x}{2}} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} = \frac{1 + \sin x}{\cos x} \]

### Step 3: Rewrite the integral
Now we can rewrite the integral:
\[ \int \tan^{-1}(\sec x + \tan x) \, dx = \int \left(\frac{\pi}{4} + \frac{x}{2}\right) \, dx \]

### Step 4: Integrate
Now we integrate:
\[ \int \left(\frac{\pi}{4} + \frac{x}{2}\right) \, dx = \frac{\pi}{4} x + \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{\pi}{4} x + \frac{x^2}{4} + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \tan^{-1}(\sec x + \tan x) \, dx = \frac{\pi}{4} x + \frac{x^2}{4} + C \]