`int sin^(-1)((2tanx )/(1+tan^(2)x))dx = ?`

Options

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

Explanation

To solve the integral

\[ \int \sin^{-1}\left(\frac{2\tan x}{1+\tan^2 x}\right) dx, \]

we can use the identity for the sine of double angles.

### Step 1: Recognize the identity
We know that

\[ \sin(2x) = \frac{2\tan x}{1 + \tan^2 x}. \]

Thus, we can rewrite the integral as:

\[ \int \sin^{-1}(\sin(2x)) \, dx. \]

### Step 2: Simplify the integral
Since \(\sin^{-1}(\sin(2x))\) is equal to \(2x\) (for \(x\) in the principal range of \(\sin^{-1}\)), we can simplify our integral:

\[ \int \sin^{-1}(\sin(2x)) \, dx = \int 2x \, dx. \]

### Step 3: Integrate
Now, we can integrate \(2x\):

\[ \int 2x \, dx = 2 \cdot \frac{x^2}{2} + C = x^2 + C. \]

### Final Answer
Thus, the integral evaluates to:

\[ \int \sin^{-1}\left(\frac{2\tan x}{1+\tan^2 x}\right) dx = x^2 + C. \]