To solve the integral \(\int \frac{(\arctan(x))^2}{1+x^2} \, dx\), let's proceed with substitution.

### Step 1: Substitution
Let:
\[
u = \arctan(x)
\]
Then, the derivative of \(u\) with respect to \(x\) is:
\[
du = \frac{1}{1+x^2} \, dx
\]
This simplifies the integral to:
\[
\int u^2 \, du
\]

### Step 2: Solve the Integral
The integral of \(u^2\) with respect to \(u\) is:
\[
\int u^2 \, du = \frac{u^3}{3} + C
\]
where \(C\) is the constant of integration.

### Step 3: Back-Substitute
Now, substitute \(u = \arctan(x)\) back into the expression:
\[
\frac{(\arctan(x))^3}{3} + C
\]

### Final Answer:
The solution to the integral is:
\[
\int \frac{(\arctan(x))^2}{1+x^2} \, dx = \frac{(\arctan(x))^3}{3} + C
\]

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:
1. How would you evaluate \(\int \frac{\arctan(x)}{1+x^2} \, dx\)?
2. What is the integral of \(\arctan(x)\) itself with respect to \(x\)?
3. Can you solve \(\int \frac{1}{1+x^2} \, dx\) and relate it to arctan(x)?
4. How does substitution work when dealing with more complex integrals?
5. What happens if the integrand is \(\frac{(\arctan(x))^n}{1+x^2}\) for a general \(n\)?

**Tip:** When dealing with integrals involving inverse trigonometric functions, substitution is often a useful method to simplify the expression.

Explore the step-by-step solution for evaluating the integral ∫(arctan(x))^2/(1+x^2) using the substitution method. This problem highlights key concepts in calculus, including integration and inverse trigonometric functions, and is suitable for undergraduate students.

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