`int (tanx sec^(2)x)/((1-tan^(2)x))dx.`

Answer

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

### Step 1: Substitute \( \tan x \)
Let \( t = \tan x \). Then, the derivative of \( t \) is:
\[ dt = \sec^2 x \, dx \]
This means that \( dx = \frac{dt}{\sec^2 x} \).

### Step 2: Rewrite the integral
Substituting \( t \) into the integral, we have:
\[ \int \frac{t \sec^2 x}{1 - t^2} \cdot \frac{dt}{\sec^2 x} = \int \frac{t}{1 - t^2} \, dt \]

### Step 3: Simplify the integral
Now we need to integrate:
\[ \int \frac{t}{1 - t^2} \, dt \]
We can simplify this by using the substitution \( u = 1 - t^2 \), which gives us \( du = -2t \, dt \) or \( dt = -\frac{du}{2t} \).

### Step 4: Substitute and integrate
Thus, we can rewrite the integral:
\[ \int \frac{t}{u} \left(-\frac{du}{2t}\right) = -\frac{1}{2} \int \frac{1}{u} \, du \]
The integral of \( \frac{1}{u} \) is:
\[ -\frac{1}{2} \ln |u| + C \]

### Step 5: Back substitute for \( u \)
Now, substituting back for \( u = 1 - t^2 \):
\[ -\frac{1}{2} \ln |1 - t^2| + C \]

### Step 6: Back substitute for \( t \)
Finally, substituting back for \( t = \tan x \):
\[ -\frac{1}{2} \ln |1 - \tan^2 x| + C \]

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