`intsqrt(1+2tanx(secx+tanx))dx=`

Options

A. `-logabs(1-sinx)+c`
B. `logabs(secx+tanx)+c`
C. `logabs(secx(cosecx-cotx))+c`
D. `(sec^(2)x+tan^(2)x)/(2sqrt(1+2tanx(secx+tanx)))+c`

Answer

To solve the integral \(\int \sqrt{1 + 2 \tan x (\sec x + \tan x)} \, dx\), we will follow these steps:

### Step 1: Simplify the expression inside the square root
We start by rewriting \(1\) in terms of \(\sec^2 x\) and \(\tan^2 x\):
\[ 1 = \sec^2 x - \tan^2 x \]
Thus, we can rewrite the integral as:
\[ \int \sqrt{\sec^2 x - \tan^2 x + 2 \tan x (\sec x + \tan x)} \, dx \]

### Step 2: Expand the expression
Next, we expand \(2 \tan x (\sec x + \tan x)\):
\[ 2 \tan x (\sec x + \tan x) = 2 \tan x \sec x + 2 \tan^2 x \]
Now substituting this back, we have:
\[ \sqrt{\sec^2 x - \tan^2 x + 2 \tan x \sec x + 2 \tan^2 x} \]
This simplifies to:
\[ \sqrt{\sec^2 x + \tan^2 x + 2 \tan x \sec x} \]

### Step 3: Recognize a perfect square
The expression inside the square root can be recognized as a perfect square:
\[ \sec^2 x + 2 \tan x \sec x + \tan^2 x = (\sec x + \tan x)^2 \]
Thus, we can rewrite the integral as:
\[ \int \sqrt{(\sec x + \tan x)^2} \, dx \]
Since \(\sec x + \tan x\) is positive in the domain we are considering, we can simplify this to:
\[ \int (\sec x + \tan x) \, dx \]

### Step 4: Integrate
The integral of \(\sec x + \tan x\) is a standard result:
\[ \int (\sec x + \tan x) \, dx = \ln |\sec x + \tan x| + C \]

### Final Answer
Thus, the final answer to the integral is:
\[ \ln |\sec x + \tan x| + C \]

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