`int (sin (2tan^(-1)x))/((1+x^(2)))dx.`

Answer

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

### Step 1: Substitution
Let \( t = \tan^{-1} x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ \frac{dt}{dx} = \frac{1}{1 + x^2} \]
This implies that:
\[ dx = (1 + x^2) \, dt \]
Thus, we can rewrite the integral as:
\[ I = \int \sin(2t) \, dt \]

### Step 2: Simplifying the Integral
Using the double angle formula for sine, we have:
\[ \sin(2t) = 2 \sin(t) \cos(t) \]
Thus, the integral becomes:
\[ I = \int 2 \sin(t) \cos(t) \, dt \]

### Step 3: Integrating
The integral of \( 2 \sin(t) \cos(t) \) can be simplified further:
\[ I = \int \sin(2t) \, dt \]
Now, we can integrate:
\[ I = -\frac{1}{2} \cos(2t) + C \]

### Step 4: Back Substitution
Now, we need to substitute back for \( t \):
\[ t = \tan^{-1} x \]
Thus, \( 2t = 2 \tan^{-1} x \). Therefore, we have:
\[ I = -\frac{1}{2} \cos(2 \tan^{-1} x) + C \]

### Final Result
The final answer for the integral is:
\[ I = -\frac{1}{2} \cos(2 \tan^{-1} x) + C \]