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

Options

A. `logtan^(2)x+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

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

we will follow these steps:

### Step 1: Simplify the integrand
We know from trigonometric identities that

\[ 1 + \tan^2 x = \sec^2 x. \]

Thus, we can rewrite the integrand as:

\[ \frac{(1+\sqrt{\tan x}) \sec^2 x}{2 \tan x}. \]

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

\[ dt = \sec^2 x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\sec^2 x}. \]

Substituting \( t \) into the integral gives:

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

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

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

### Step 4: Integrate each part
1. For the first integral:

\[ \int \frac{1}{2t} \, dt = \frac{1}{2} \ln |t| + C_1. \]

2. For the second integral:

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

### Step 5: Combine the results
Combining both results, we have:

\[ \frac{1}{2} \ln |t| + \sqrt{t} + C, \]

where \( C = C_1 + C_2 \).

### Step 6: Substitute back for \( t \)
Recall that \( t = \tan x \). Therefore, substituting back gives:

\[ \frac{1}{2} \ln |\tan x| + \sqrt{\tan x} + C. \]

### Final Result
Thus, the integral evaluates to:

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

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