`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: Rewrite the integral
Start by rewriting the integral in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus, we can express \( 1 + \tan x \) and \( 1 - \tan x \) as:
\[ 1 + \tan x = 1 + \frac{\sin x}{\cos x} = \frac{\cos x + \sin x}{\cos x} \]
\[ 1 - \tan x = 1 - \frac{\sin x}{\cos x} = \frac{\cos x - \sin x}{\cos x} \]
Now substituting these into the integral:
\[ \int \frac{1 + \tan x}{1 - \tan x} \, dx = \int \frac{\frac{\cos x + \sin x}{\cos x}}{\frac{\cos x - \sin x}{\cos x}} \, dx = \int \frac{\cos x + \sin x}{\cos x - \sin x} \, dx \]

### Step 2: Simplify the integral
Now, we can simplify the integral:
\[ \int \frac{\cos x + \sin x}{\cos x - \sin x} \, dx \]

### Step 3: Use substitution
Let \( t = \cos x - \sin x \). Then, we differentiate:
\[ dt = (-\sin x - \cos x) \, dx \quad \Rightarrow \quad dx = \frac{dt}{-\sin x - \cos x} \]
Now we need to express \( \cos x + \sin x \) in terms of \( t \):
\[ \cos x + \sin x = \sqrt{(\cos x - \sin x)^2 + 2\sin x \cos x} = \sqrt{t^2 + 2\sin x \cos x} \]
However, we can also notice that:
\[ \sin x + \cos x = -dt \]
Thus, we can write:
\[ \int \frac{\cos x + \sin x}{\cos x - \sin x} \, dx = \int \frac{-dt}{t} \]

### Step 4: Integrate
Now, we can integrate:
\[ \int \frac{-dt}{t} = -\ln |t| + C \]

### Step 5: Substitute back
Substituting back \( t = \cos x - \sin x \):
\[ -\ln |\cos x - \sin x| + C \]

### Step 6: Final answer
Thus, the final answer is:
\[ \int \frac{1 + \tan x}{1 - \tan x} \, dx = -\ln |\cos x - \sin x| + C \]