`inttan^(-1)sqrt(x)\ dx` is equal to

Options

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

Explanation

To solve the integral \( \int \tan^{-1}(\sqrt{x}) \, dx \), we will use integration by parts. Let's denote:

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

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \tan^{-1}(\sqrt{x}) \)
- \( dv = dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now we need to find \( du \) and \( v \):
- To find \( du \), we use the derivative of \( \tan^{-1}(x) \):
\[ du = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{1}{2\sqrt{x}(1+x)} \, dx \]
- Integrating \( dv \):
\[ v = x \]

### Step 3: Apply Integration by Parts Formula
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I = x \tan^{-1}(\sqrt{x}) - \int x \cdot \frac{1}{2\sqrt{x}(1+x)} \, dx \]

### Step 4: Simplify the Integral
The integral becomes:
\[ I = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \int \frac{x}{\sqrt{x}(1+x)} \, dx \]
\[ = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{1+x} \, dx \]

### Step 5: Substitute \( x = t^2 \)
Let \( x = t^2 \), then \( dx = 2t \, dt \) and \( \sqrt{x} = t \):
\[ I = t^2 \tan^{-1}(t) - \frac{1}{2} \int \frac{t}{1+t^2} \cdot 2t \, dt \]
\[ = t^2 \tan^{-1}(t) - \int \frac{t^2}{1+t^2} \, dt \]

### Step 6: Simplify the Integral
The integral can be simplified:
\[ \int \frac{t^2}{1+t^2} \, dt = \int \left( 1 - \frac{1}{1+t^2} \right) dt = \int dt - \int \frac{1}{1+t^2} dt \]
\[ = t - \tan^{-1}(t) \]

### Step 7: Substitute Back
Now substituting back:
\[ I = t^2 \tan^{-1}(t) - \left( t - \tan^{-1}(t) \right) + C \]
Substituting \( t = \sqrt{x} \):
\[ I = x \tan^{-1}(\sqrt{x}) - \left( \sqrt{x} - \tan^{-1}(\sqrt{x}) \right) + C \]
\[ = x \tan^{-1}(\sqrt{x}) - \sqrt{x} + \tan^{-1}(\sqrt{x}) + C \]

### Final Result
Combining the terms gives:
\[ I = (x + 1) \tan^{-1}(\sqrt{x}) - \sqrt{x} + C \]