`int {1+2 tan x (tan x + sec x )}^(1//2) dx=?`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \sqrt{1 + 2 \tan x (\tan x + \sec x)} \, dx \]

### Step 2: Expand the Expression Inside the Square Root
Now, we expand the expression inside the square root:
\[ 1 + 2 \tan x (\tan x + \sec x) = 1 + 2 \tan^2 x + 2 \tan x \sec x \]

### Step 3: Substitute \(1\) with \(\sec^2 x - \tan^2 x\)
Using the identity \(1 + \tan^2 x = \sec^2 x\), we can rewrite \(1\) as:
\[ 1 = \sec^2 x - \tan^2 x \]
Thus, we have:
\[ \sec^2 x - \tan^2 x + 2 \tan^2 x + 2 \tan x \sec x = \sec^2 x + \tan^2 x + 2 \tan x \sec x \]

### Step 4: Factor the Expression
Notice that:
\[ \sec^2 x + \tan^2 x + 2 \tan x \sec x = (\sec x + \tan x)^2 \]
So we can rewrite the integral as:
\[ I = \int \sqrt{(\sec x + \tan x)^2} \, dx \]

### Step 5: Simplify the Integral
Since \(\sqrt{(\sec x + \tan x)^2} = |\sec x + \tan x|\), and for \(x\) in the appropriate range, we can drop the absolute value:
\[ I = \int (\sec x + \tan x) \, dx \]

### Step 6: Integrate
The integral of \(\sec x + \tan x\) is:
\[ \int \sec x \, dx + \int \tan x \, dx \]
The results are:
\[ \int \sec x \, dx = \ln |\sec x + \tan x| + C_1 \]
\[ \int \tan x \, dx = -\ln |\cos x| + C_2 \]
Thus, combining these results:
\[ I = \ln |\sec x + \tan x| - \ln |\cos x| + C \]

### Step 7: Combine Logarithms
Using the properties of logarithms, we can combine these:
\[ I = \ln \left( \frac{|\sec x + \tan x|}{|\cos x|} \right) + C \]
Since \(\sec x = \frac{1}{\cos x}\), we can simplify further:
\[ I = \ln \left( \frac{1 + \sin x}{\cos x} \right) + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \ln \left( \sec x + \tan x \right) + C \]