`int tan^(-1)sqrt xdx` is equal to:

Options

A. `(x+1) tan^(-1) sqrtx-sqrtx+c`
B. `xtan^(-1) sqrt x-sqrt x+c`
C. `sqrt x -x tan^(-1)sqrt x+c`
D. `sqrt x-(x+1)tan^(-1)sqrtx+c`

Answer

To solve the integral \( \int \tan^{-1}(\sqrt{x}) \, dx \), we can use integration by parts. Let's go through the steps systematically.

### Step 1: Set up the integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \tan^{-1}(\sqrt{x}) \)
- \( dv = dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{d}{dx}(\tan^{-1}(\sqrt{x})) \, dx = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{1}{2(1+x)} \, dx \]
- Integrate \( dv \):
\[ v = \int dx = x \]

### Step 3: Apply the integration by parts formula
Substituting \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula:
\[ \int \tan^{-1}(\sqrt{x}) \, dx = x \tan^{-1}(\sqrt{x}) - \int x \cdot \frac{1}{2(1+x)} \, dx \]

### Step 4: Simplify the integral
Now we need to simplify the integral:
\[ \int x \cdot \frac{1}{2(1+x)} \, dx = \frac{1}{2} \int \frac{x}{1+x} \, dx \]
We can rewrite \( \frac{x}{1+x} \) as:
\[ \frac{x}{1+x} = 1 - \frac{1}{1+x} \]
Thus,
\[ \int \frac{x}{1+x} \, dx = \int \left(1 - \frac{1}{1+x}\right) \, dx = \int 1 \, dx - \int \frac{1}{1+x} \, dx = x - \ln|1+x| \]

### Step 5: Substitute back into the equation
Now substituting back:
\[ \int \tan^{-1}(\sqrt{x}) \, dx = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \left( x - \ln|1+x| \right) + C \]
This simplifies to:
\[ \int \tan^{-1}(\sqrt{x}) \, dx = x \tan^{-1}(\sqrt{x}) - \frac{x}{2} + \frac{1}{2} \ln|1+x| + C \]

### Step 6: Final expression
Thus, the final expression for the integral is:
\[ \int \tan^{-1}(\sqrt{x}) \, dx = x \tan^{-1}(\sqrt{x}) - \frac{x}{2} + \frac{1}{2} \ln(1+x) + C \]