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

Answer

To solve the integral \( \int \tan^{-1}(x) \, dx \), we will use integration by parts. Here is the step-by-step solution:

### Step 1: Identify parts for integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \tan^{-1}(x) \) (which we will differentiate)
- \( dv = dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now we differentiate \( u \) and integrate \( dv \):
- \( du = \frac{1}{1+x^2} \, dx \)
- \( v = x \)

### Step 3: Apply the 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 simplify the integral \( \int \frac{x}{1+x^2} \, dx \). We can rewrite it as:
\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \int \frac{2x}{1+x^2} \, dx \]
Let \( t = 1 + x^2 \), then \( dt = 2x \, dx \).

### Step 5: Substitute and integrate
Substituting \( t \) into the integral:
\[ \int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dt = \ln |t| + C = \ln(1+x^2) + C \]
Thus,
\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) + C \]

### Step 6: Combine results
Now, substituting back into our integration by parts result:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \left( \frac{1}{2} \ln(1+x^2) \right) + C \]
Finally, we can write the answer as:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]

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

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