Evaluate: `int"t"a n^(-1)sqrt(x)dx`

Explanation

To evaluate the integral \( \int \tan^{-1}(\sqrt{x}) \, dx \), we will use substitution and integration by parts. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, we have:
\[ x = t^2 \quad \text{and} \quad dx = 2t \, dt \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we get:
\[ \int \tan^{-1}(\sqrt{x}) \, dx = \int \tan^{-1}(t) \cdot 2t \, dt = 2 \int t \tan^{-1}(t) \, dt \]

### Step 3: Integration by Parts
Let:
- \( u = \tan^{-1}(t) \) (which means \( du = \frac{1}{1+t^2} \, dt \))
- \( dv = t \, dt \) (which means \( v = \frac{t^2}{2} \))

Using integration by parts, \( \int u \, dv = uv - \int v \, du \):
\[ \int t \tan^{-1}(t) \, dt = \tan^{-1}(t) \cdot \frac{t^2}{2} - \int \frac{t^2}{2} \cdot \frac{1}{1+t^2} \, dt \]

### Step 4: Simplifying the Integral
The integral simplifies to:
\[ \int t \tan^{-1}(t) \, dt = \frac{t^2}{2} \tan^{-1}(t) - \frac{1}{2} \int \frac{t^2}{1+t^2} \, dt \]
Now, simplify \( \frac{t^2}{1+t^2} \):
\[ \frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2} \]
Thus,
\[ \int \frac{t^2}{1+t^2} \, dt = \int 1 \, dt - \int \frac{1}{1+t^2} \, dt = t - \tan^{-1}(t) \]

### Step 5: Substitute Back
Now substituting back into the integral:
\[ \int t \tan^{-1}(t) \, dt = \frac{t^2}{2} \tan^{-1}(t) - \frac{1}{2} \left( t - \tan^{-1}(t) \right) \]
This gives us:
\[ = \frac{t^2}{2} \tan^{-1}(t) - \frac{t}{2} + \frac{1}{2} \tan^{-1}(t) \]

### Step 6: Combine Terms
Combining the terms:
\[ = \left( \frac{t^2}{2} + \frac{1}{2} \right) \tan^{-1}(t) - \frac{t}{2} \]

### Step 7: Substitute \( t = \sqrt{x} \)
Substituting \( t = \sqrt{x} \) back into the equation:
\[ = \left( \frac{x}{2} + \frac{1}{2} \right) \tan^{-1}(\sqrt{x}) - \frac{\sqrt{x}}{2} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \tan^{-1}(\sqrt{x}) \, dx = \left( \frac{x+1}{2} \right) \tan^{-1}(\sqrt{x}) - \frac{\sqrt{x}}{2} + C \]