Compute the integrals :
`int_(0)^(1) " arc tan " sqrt(x) dx`

Answer

To compute the integral \( \int_{0}^{1} \tan^{-1}(\sqrt{x}) \, dx \), we will use integration by parts. The formula for integration by parts is:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = \tan^{-1}(\sqrt{x}) \) (which we will differentiate)
- \( dv = dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and Integrate \( dv \)

Now we need to find \( du \) and \( v \):
- To find \( du \), we differentiate \( u \):

\[ du = \frac{d}{dx}(\tan^{-1}(\sqrt{x})) \, dx = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{1}{2(1+x)\sqrt{x}} \, dx \]

- To find \( v \), we integrate \( dv \):

\[ v = \int dx = x \]

### Step 3: Apply Integration by Parts Formula

Now we can apply the integration by parts formula:

\[ \int_{0}^{1} \tan^{-1}(\sqrt{x}) \, dx = \left[ x \tan^{-1}(\sqrt{x}) \right]_{0}^{1} - \int_{0}^{1} x \cdot \frac{1}{2(1+x)\sqrt{x}} \, dx \]

### Step 4: Evaluate the Boundary Terms

Evaluate \( \left[ x \tan^{-1}(\sqrt{x}) \right]_{0}^{1} \):

- At \( x = 1 \):

\[ 1 \cdot \tan^{-1}(\sqrt{1}) = 1 \cdot \tan^{-1}(1) = 1 \cdot \frac{\pi}{4} = \frac{\pi}{4} \]

- At \( x = 0 \):

\[ 0 \cdot \tan^{-1}(\sqrt{0}) = 0 \cdot \tan^{-1}(0) = 0 \]

Thus, the boundary term evaluates to:

\[ \left[ x \tan^{-1}(\sqrt{x}) \right]_{0}^{1} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Step 5: Simplify the Remaining Integral

Now we need to simplify the integral:

\[ -\int_{0}^{1} x \cdot \frac{1}{2(1+x)\sqrt{x}} \, dx = -\frac{1}{2} \int_{0}^{1} \frac{x}{(1+x)\sqrt{x}} \, dx \]

This can be rewritten as:

\[ -\frac{1}{2} \int_{0}^{1} \frac{\sqrt{x}}{1+x} \, dx \]

### Step 6: Evaluate the Integral

To evaluate \( \int_{0}^{1} \frac{\sqrt{x}}{1+x} \, dx \), we can use the substitution \( x = t^2 \), which gives \( dx = 2t \, dt \):

\[ \int_{0}^{1} \frac{\sqrt{x}}{1+x} \, dx = \int_{0}^{1} \frac{t}{1+t^2} \cdot 2t \, dt = 2 \int_{0}^{1} \frac{t^2}{1+t^2} \, dt \]

This integral can be split as:

\[ 2 \int_{0}^{1} \left( 1 - \frac{1}{1+t^2} \right) dt = 2 \left[ t - \tan^{-1}(t) \right]_{0}^{1} = 2 \left( 1 - \frac{\pi}{4} \right) = 2 - \frac{\pi}{2} \]

### Step 7: Combine Results

Now substituting back, we have:

\[ -\frac{1}{2} \left( 2 - \frac{\pi}{2} \right) = -1 + \frac{\pi}{4} \]

### Final Result

Combining everything:

\[ \int_{0}^{1} \tan^{-1}(\sqrt{x}) \, dx = \frac{\pi}{4} - \left( -1 + \frac{\pi}{4} \right) = 1 \]

Thus, the final answer is:

\[ \int_{0}^{1} \tan^{-1}(\sqrt{x}) \, dx = 1 \]