`int(1+tan^(2)x)/(1+tanx)dx=`

Options

A. `sinx+cosx+c`
B. `cosx-sinx+c`
C. `log(1+tanx)+c`
D. `log(cosx+sinx)+c`

Answer

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

### Step 1: Simplify the integrand
We know that \( 1 + \tan^2 x = \sec^2 x \). Therefore, we can rewrite the integral as:
\[ \int \frac{\sec^2 x}{1 + \tan x} \, dx \]

### Step 2: Use substitution
Let \( t = 1 + \tan x \). Then, we need to find \( dt \):
\[ dt = \sec^2 x \, dx \]
This means that \( dx = \frac{dt}{\sec^2 x} \).

### Step 3: Substitute in the integral
Substituting \( t \) and \( dt \) into the integral gives us:
\[ \int \frac{\sec^2 x}{t} \cdot \frac{dt}{\sec^2 x} = \int \frac{1}{t} \, dt \]

### Step 4: Integrate
The integral of \( \frac{1}{t} \) is:
\[ \ln |t| + C \]

### Step 5: Substitute back
Now we substitute back \( t = 1 + \tan x \):
\[ \ln |1 + \tan x| + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{1 + \tan^2 x}{1 + \tan x} \, dx = \ln |1 + \tan x| + C \]

---