To solve the integral \( I = \int \tan^{-1}(\sqrt{x}) \, dx \), we will use integration by parts. The formula for integration by parts is given by:
\[
\int u \, dv = uv - \int v \, du
\]
### 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 differentiate \( u \):
\[
du = \frac{d}{dx} \tan^{-1}(\sqrt{x}) \, dx = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{1}{2\sqrt{x}(1+x)} \, dx
\]
- To find \( v \), we integrate \( dv \):
\[
v = \int dx = x
\]
### Step 3: Apply the integration by parts formula
Now we can apply the integration by parts formula:
\[
I = uv - \int v \, du
\]
Substituting the values we found:
\[
I = \tan^{-1}(\sqrt{x}) \cdot x - \int x \cdot \frac{1}{2\sqrt{x}(1+x)} \, dx
\]
### Step 4: Simplify the integral
Now we simplify the integral:
\[
I = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \int \frac{x}{\sqrt{x}(1+x)} \, dx = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{1+x} \, dx
\]
### Step 5: Substitute \( \sqrt{x} = t \)
Let \( \sqrt{x} = t \), then \( x = t^2 \) and \( dx = 2t \, dt \). Substitute these into the integral:
\[
I = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \int \frac{t}{1+t^2} \cdot 2t \, dt
\]
This simplifies to:
\[
I = x \tan^{-1}(\sqrt{x}) - \int \frac{t^2}{1+t^2} \, dt
\]
### Step 6: Split the integral
Now we can split the integral:
\[
\int \frac{t^2}{1+t^2} \, dt = \int \left( 1 - \frac{1}{1+t^2} \right) dt = \int dt - \int \frac{1}{1+t^2} dt
\]
### Step 7: Evaluate the integrals
Evaluating these integrals gives:
\[
\int dt = t \quad \text{and} \quad \int \frac{1}{1+t^2} dt = \tan^{-1}(t)
\]
Thus,
\[
\int \frac{t^2}{1+t^2} \, dt = t - \tan^{-1}(t)
\]
### Step 8: Substitute back
Substituting back into our expression for \( I \):
\[
I = x \tan^{-1}(\sqrt{x}) - \left( \sqrt{x} - \tan^{-1}(\sqrt{x}) \right) + C
\]
### Step 9: Combine terms
Combining the terms gives:
\[
I = x \tan^{-1}(\sqrt{x}) - \sqrt{x} + \tan^{-1}(\sqrt{x}) + C
\]
### Final Result
Thus, the final result is:
\[
I = \tan^{-1}(\sqrt{x})(x + 1) - \sqrt{x} + C
\]
