`int((1+sqrt(tanx))(1+tan^(2)x))/(2tanx)dx` equals to

Options

A. `logsqrt(tanx)+sqrt(tanx)+c`
B. `logtan^(2)x+(1)/(2sqrt(tanx))+c`
C. `log|tanx|+2sqrt(tanx)+c`
D. none of these

Answer

To solve the integral

\[ I = \int \frac{(1 + \sqrt{\tan x})(1 + \tan^2 x)}{2 \tan x} \, dx, \]

we will 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:

\[ I = \int \frac{(1 + \sqrt{\tan x}) \sec^2 x}{2 \tan x} \, dx. \]

### Step 2: Substitute \( \tan x = t \)
Let \( t = \tan x \). Then, the derivative is:

\[ \frac{dt}{dx} = \sec^2 x \implies dx = \frac{dt}{\sec^2 x}. \]

Substituting \(dx\) into the integral gives:

\[ I = \int \frac{(1 + \sqrt{t}) \sec^2 x}{2t} \cdot \frac{dt}{\sec^2 x} = \int \frac{(1 + \sqrt{t})}{2t} \, dt. \]

### Step 3: Split the integral
Now we can split the integral into two parts:

\[ I = \frac{1}{2} \int \frac{1}{t} \, dt + \frac{1}{2} \int \frac{\sqrt{t}}{t} \, dt = \frac{1}{2} \int \frac{1}{t} \, dt + \frac{1}{2} \int t^{-\frac{1}{2}} \, dt. \]

### Step 4: Integrate each term
1. The first integral:

\[ \int \frac{1}{t} \, dt = \log |t| + C_1. \]

2. The second integral:

\[ \int t^{-\frac{1}{2}} \, dt = 2t^{\frac{1}{2}} + C_2. \]

### Step 5: Combine the results
Putting it all together, we have:

\[ I = \frac{1}{2} \left( \log |t| + 2\sqrt{t} \right) + C. \]

### Step 6: Substitute back \(t = \tan x\)
Now, substituting back \(t = \tan x\):

\[ I = \frac{1}{2} \left( \log |\tan x| + 2\sqrt{\tan x} \right) + C. \]

### Step 7: Final simplification
This simplifies to:

\[ I = \frac{1}{2} \log |\tan x| + \sqrt{\tan x} + C. \]

### Final Answer
Thus, the final answer is:

\[ I = \frac{1}{2} \log |\tan x| + \sqrt{\tan x} + C. \]

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