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

Options

A. `(1)/(2)log|sec2x+tan2x|+c`
B. `(-1)/(2)log|sec2x+tan2x|+c`
C. `2log|sec2x+tan2x|+c`
D. `-2log|sec2x+tan2x|+c`

Answer

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

### Step 1: Rewrite the integral using trigonometric identities
We know that \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \). Therefore, we can rewrite the integral as:
\[ I = \int \frac{1 + \frac{\sin^2 x}{\cos^2 x}}{1 - \frac{\sin^2 x}{\cos^2 x}} \, dx \]

### Step 2: Simplify the expression
Combine the terms in the numerator and denominator:
\[ I = \int \frac{\frac{\cos^2 x + \sin^2 x}{\cos^2 x}}{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}} \, dx \]
This simplifies to:
\[ I = \int \frac{\cos^2 x + \sin^2 x}{\cos^2 x - \sin^2 x} \, dx \]
Using the identity \( \cos^2 x + \sin^2 x = 1 \), we have:
\[ I = \int \frac{1}{\cos^2 x - \sin^2 x} \, dx \]

### Step 3: Use the identity for cosine
Recall that \( \cos^2 x - \sin^2 x = \cos 2x \). Thus, we can rewrite the integral as:
\[ I = \int \frac{1}{\cos 2x} \, dx \]

### Step 4: Rewrite in terms of secant
This can be expressed as:
\[ I = \int \sec 2x \, dx \]

### Step 5: Integrate using the formula for secant
The integral of \( \sec \theta \) is given by:
\[ \int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| + C \]
Applying this to our integral:
\[ I = \frac{1}{2} \ln |\sec 2x + \tan 2x| + C \]

### Final Answer
Thus, the final result is:
\[ I = \frac{1}{2} \ln |\sec 2x + \tan 2x| + C \]