`int_(-pi//4)^0 (1+tanx)/(1-tanx) dx`

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx, \]

we can start by simplifying the integrand.

### Step 1: Rewrite the integrand
We can express the integrand in terms of a simpler function. Notice that:

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

This is useful because \(1 - \tan^2 x = \cos^2 x\) when we apply the identity \(1 + \tan^2 x = \sec^2 x\).

### Step 2: Use the property of definite integrals
We can also use the property of definite integrals. Specifically, we can use the substitution \(x = -t\), which gives us:

\[ dx = -dt. \]

Changing the limits accordingly, when \(x = -\frac{\pi}{4}\), \(t = \frac{\pi}{4}\) and when \(x = 0\), \(t = 0\). Thus, we have:

\[ I = \int_{\frac{\pi}{4}}^{0} \frac{1 + \tan(-t)}{1 - \tan(-t)} (-dt) = \int_{0}^{\frac{\pi}{4}} \frac{1 - \tan t}{1 + \tan t} dt. \]

### Step 3: Combine the integrals
Now we can combine the two integrals:

\[ I = \int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx + \int_{0}^{\frac{\pi}{4}} \frac{1 - \tan t}{1 + \tan t} \, dt. \]

Notice that:

\[ \frac{1 - \tan t}{1 + \tan t} = \frac{1 + \tan(-t)}{1 - \tan(-t)}. \]

Thus, we can write:

\[ I = \int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx + \int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx = 2I. \]

### Step 4: Solve for I
This leads us to:

\[ 2I = \int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx + \int_{0}^{\frac{\pi}{4}} \frac{1 - \tan t}{1 + \tan t} \, dt = \int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx + \int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx. \]

### Step 5: Calculate the integral
Now we can evaluate the integral:

\[ I = \frac{1}{2} \int_{-\frac{\pi}{4}}^{0} \frac{1 + \tan x}{1 - \tan x} \, dx. \]

To evaluate this integral, we can use the substitution \(u = \tan x\), which gives \(du = \sec^2 x \, dx\). The limits change from \(x = -\frac{\pi}{4}\) to \(x = 0\) which translates to \(u = -1\) to \(u = 0\).

### Final Result
After evaluating the integral, we find:

\[ I = -\frac{1}{2} \log(2). \]