To solve the integral \( \int \tan^{-1}(\sqrt{x}) \, dx \), we will follow these steps:
### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, \( x = t^2 \) and differentiating gives:
\[
dx = 2t \, dt
\]
### Step 2: Rewrite the Integral
Now, substitute \( t \) and \( dx \) into the integral:
\[
\int \tan^{-1}(\sqrt{x}) \, dx = \int \tan^{-1}(t) \cdot (2t \, dt) = 2 \int t \tan^{-1}(t) \, dt
\]
### Step 3: Integration by Parts
We will use integration by parts, where we let:
- \( u = \tan^{-1}(t) \) (first function)
- \( dv = t \, dt \) (second function)
Then, we differentiate and integrate:
- \( du = \frac{1}{1+t^2} \, dt \)
- \( v = \frac{t^2}{2} \)
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[
\int t \tan^{-1}(t) \, dt = \tan^{-1}(t) \cdot \frac{t^2}{2} - \int \frac{t^2}{2} \cdot \frac{1}{1+t^2} \, dt
\]
### Step 4: Simplifying the Integral
Now we simplify the remaining integral:
\[
\int \frac{t^2}{2(1+t^2)} \, dt = \frac{1}{2} \int \left(1 - \frac{1}{1+t^2}\right) \, dt = \frac{1}{2} \left( t - \tan^{-1}(t) \right) + C
\]
### Step 5: Combine Results
Now substituting back into our integration by parts result:
\[
\int t \tan^{-1}(t) \, dt = \frac{t^2}{2} \tan^{-1}(t) - \frac{1}{2} \left( t - \tan^{-1}(t) \right) + C
\]
\[
= \frac{t^2}{2} \tan^{-1}(t) - \frac{t}{2} + \frac{1}{2} \tan^{-1}(t) + C
\]
### Step 6: Substitute Back to \( x \)
Now, replace \( t \) back with \( \sqrt{x} \):
\[
= \frac{(\sqrt{x})^2}{2} \tan^{-1}(\sqrt{x}) - \frac{\sqrt{x}}{2} + \frac{1}{2} \tan^{-1}(\sqrt{x}) + C
\]
\[
= \frac{x}{2} \tan^{-1}(\sqrt{x}) - \frac{\sqrt{x}}{2} + \frac{1}{2} \tan^{-1}(\sqrt{x}) + C
\]
### Final Result
Combining the terms gives us:
\[
\int \tan^{-1}(\sqrt{x}) \, dx = \frac{x}{2} \tan^{-1}(\sqrt{x}) - \frac{\sqrt{x}}{2} + C
\]
