`inttan^(-1)((2tanx)/(1-tan^(2)x))dx=`

Options

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

Answer

To solve the integral \( I = \int \tan^{-1}\left(\frac{2\tan x}{1 - \tan^2 x}\right) dx \), we can follow these steps:

### Step 1: Recognize the formula
We know that:
\[ \tan(2x) = \frac{2\tan x}{1 - \tan^2 x} \]
This means that:
\[ \frac{2\tan x}{1 - \tan^2 x} = \tan(2x) \]
Thus, we can rewrite the integral as:
\[ I = \int \tan^{-1}(\tan(2x)) \, dx \]

### Step 2: Simplify the integral
Since \( \tan^{-1}(\tan(\theta)) = \theta \) for \( \theta \) in the principal range of \( \tan^{-1} \), we have:
\[ I = \int 2x \, dx \]

### Step 3: Integrate
Now, we can integrate \( 2x \):
\[ I = 2 \int x \, dx \]
The integral of \( x \) is:
\[ \int x \, dx = \frac{x^2}{2} \]
Thus,
\[ I = 2 \cdot \frac{x^2}{2} + C \]

### Step 4: Simplify the result
The \( 2 \) in the numerator and denominator cancels out:
\[ I = x^2 + C \]

### Final Result
Therefore, the final answer is:
\[ I = x^2 + C \]

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