`int((1+tanx)/(1-tanx))dx`

Explanation

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

### Step 1: Simplify the integrand
We start with the expression:
\[ \frac{1 + \tan x}{1 - \tan x} \]
This can be rewritten using the identity for tangent:
\[ \frac{1 + \tan x}{1 - \tan x} = \tan\left(\frac{\pi}{4} + x\right) \]
This identity helps us express the integrand in a more manageable form.

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

### Step 3: Integrate
The integral of \(\tan u\) is:
\[ \int \tan u \, du = -\log|\cos u| + C \]
Applying this to our integral:
\[ \int \tan\left(\frac{\pi}{4} + x\right) \, dx = -\log|\cos\left(\frac{\pi}{4} + x\right)| + C \]

### Step 4: Final expression
Thus, we can express the final result as:
\[ -\log|\cos\left(\frac{\pi}{4} + x\right)| + C \]
Alternatively, using the secant function:
\[ \log|\sec\left(\frac{\pi}{4} + x\right)| + C \]

### Final Answer
The integral evaluates to:
\[ \int \frac{1 + \tan x}{1 - \tan x} \, dx = -\log|\cos\left(\frac{\pi}{4} + x\right)| + C \quad \text{or} \quad \log|\sec\left(\frac{\pi}{4} + x\right)| + C \]