`inttan^(-1)xdx=xtan^(-1)x+f(x)+C, " find "f(x)`.

Answer

To solve the integral \( \int \tan^{-1}(x) \, dx \) and find the function \( f(x) \) in the equation

\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) + f(x) + C, \]

we will use integration by parts.

### Step-by-Step Solution:

1. **Choose Functions for Integration by Parts**:
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du. \]
Let:
- \( u = \tan^{-1}(x) \) (first function)
- \( dv = dx \) (second function)

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

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

4. **Simplify the Remaining Integral**:
The remaining integral is:
\[ \int \frac{x}{1 + x^2} \, dx. \]
We can use the substitution \( t = 1 + x^2 \), which gives \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \). Thus:
\[ \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \int \frac{1}{t} \, dt = \frac{1}{2} \ln |t| + C = \frac{1}{2} \ln(1 + x^2) + C. \]

5. **Combine the Results**:
Substitute back into the integration by parts result:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1 + x^2) + C. \]

6. **Identify \( f(x) \)**:
From the original equation:
\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) + f(x) + C, \]
we can equate:
\[ f(x) = -\frac{1}{2} \ln(1 + x^2). \]

### Final Answer:
Thus, the function \( f(x) \) is:
\[ f(x) = -\frac{1}{2} \ln(1 + x^2). \]