`intsqrt[tanx].dx`

Explanation

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

### Step 1: Substitution
Let \( t = \sqrt{\tan x} \). Then, we have:
\[ \tan x = t^2 \]
Differentiating both sides with respect to \( x \):
\[ \sec^2 x \, dx = 2t \, dt \]
Thus, we can express \( dx \) as:
\[ dx = \frac{2t}{\sec^2 x} \, dt \]

### Step 2: Express \( \sec^2 x \) in terms of \( t \)
Using the identity \( \sec^2 x = 1 + \tan^2 x \):
\[ \sec^2 x = 1 + t^4 \]
So, substituting this into the expression for \( dx \):
\[ dx = \frac{2t}{1 + t^4} \, dt \]

### Step 3: Substitute into the integral
Now substituting \( dx \) into the integral:
\[ I = \int \sqrt{\tan x} \, dx = \int t \cdot \frac{2t}{1 + t^4} \, dt = \int \frac{2t^2}{1 + t^4} \, dt \]

### Step 4: Split the integral
We can rewrite \( 2t^2 \) as \( (t^2 + 1) + (t^2 - 1) \):
\[ I = \int \frac{t^2 + 1}{1 + t^4} \, dt + \int \frac{t^2 - 1}{1 + t^4} \, dt \]

### Step 5: Simplify the integrals
Now we can separate the two integrals:
\[ I = \int \frac{t^2 + 1}{1 + t^4} \, dt + \int \frac{t^2 - 1}{1 + t^4} \, dt \]

### Step 6: Use partial fractions
For the first integral:
\[ \int \frac{t^2 + 1}{1 + t^4} \, dt = \int \frac{1}{2} \left( \frac{1}{t^2 + 1} + \frac{1}{1 + t^4} \right) dt \]
For the second integral:
\[ \int \frac{t^2 - 1}{1 + t^4} \, dt = \int \frac{1}{2} \left( \frac{1}{t^2 + 1} - \frac{1}{1 + t^4} \right) dt \]

### Step 7: Integrate each part
The integrals can be solved using standard forms:
1. \( \int \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) \)
2. The integral of \( \frac{1}{1 + t^4} \) can be solved using trigonometric identities or tables.

### Step 8: Back substitution
After integrating, substitute back \( t = \sqrt{\tan x} \) to express the final answer in terms of \( x \).

### Final Answer
The final expression will involve \( \tan^{-1}(\sqrt{\tan x}) \) and logarithmic terms derived from the integration process.