`int (tan^(-1)x)dx`

Answer

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

### Step 1: Identify Functions for Integration by Parts
We will use the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Here, we choose:
- \( u = \tan^{-1}(x) \) (first function)
- \( dv = dx \) (second function)

### Step 2: Differentiate and Integrate
Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{1 + x^2} \, dx \]
- Integrate \( dv \):
\[ v = x \]

### Step 3: Apply Integration by Parts Formula
Substituting into the integration by parts formula:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \int x \cdot \frac{1}{1 + x^2} \, dx \]

### Step 4: Simplify the Remaining Integral
Now we need to evaluate the integral:
\[ \int \frac{x}{1 + x^2} \, dx \]
To solve this, we can use a substitution:
Let \( t = 1 + x^2 \), then \( dt = 2x \, dx \) or \( x \, dx = \frac{dt}{2} \).

### Step 5: Change of Variables
Substituting in the integral:
\[ \int \frac{x}{1 + x^2} \, dx = \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{1}{t} \, dt \]
This evaluates to:
\[ \frac{1}{2} \ln |t| + C = \frac{1}{2} \ln |1 + x^2| + C \]

### Step 6: Combine Results
Now we substitute this back into our expression:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \left( \frac{1}{2} \ln(1 + x^2) \right) + C \]
Thus, the final result is:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1 + x^2) + C \]

### Final Answer
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1 + x^2) + C \]