To solve the integral of \( \tan^{-1}(x) \), or \( \int \tan^{-1}(x) \, dx \), we can use integration by parts. Here's the step-by-step solution:

### Step 1: Apply integration by parts formula
The formula for integration by parts is:

\[
\int u \, dv = uv - \int v \, du
\]

Let:
- \( u = \tan^{-1}(x) \), so \( du = \frac{1}{1+x^2} \, dx \)
- \( dv = dx \), so \( v = x \)

### Step 2: Substitute into the formula
Now, applying the integration by parts formula:

\[
\int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \int \frac{x}{1+x^2} \, dx
\]

### Step 3: Simplify the remaining integral
The remaining integral is:

\[
\int \frac{x}{1+x^2} \, dx
\]

This can be solved by using the substitution \( u = 1 + x^2 \), so \( du = 2x \, dx \). Therefore:

\[
\int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) + C
\]

### Step 4: Final answer
Now, substitute back into the original expression:

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

Thus, the integral of \( \tan^{-1}(x) \) is:

\[
\boxed{x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C}
\]

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Would you like more details or have any questions? Here are some related questions to expand on this topic:

1. What is integration by parts, and how is it derived?
2. Can you explain why the substitution method was used for the second integral?
3. How would you integrate other inverse trigonometric functions like \( \sin^{-1}(x) \) or \( \cos^{-1}(x) \)?
4. Can this result be verified by differentiation?
5. What are some common applications of inverse trigonometric integrals in real-world problems?

**Tip**: Always check your integral by differentiating the result to ensure correctness!

Solve the integral of tan^-1(x) using integration by parts. This detailed step-by-step solution covers integration techniques including substitution. Suitable for advanced high school and college students.

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