Interesting. How do you solve it? I don't see an obvious u-substitution.

You're right that a straightforward \( u \)-substitution isn't immediately obvious. Instead, this problem is more effectively solved using **integration by parts**.

### Step-by-Step Solution:

1. **Set up the integral**:
   \[
   I = \int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \, dx
   \]

2. **Choose functions for integration by parts**:  
   Let:
   \[
   u = \ln(1+x) \quad \text{and} \quad dv = \frac{1}{1+x^2} \, dx
   \]
   Then:
   \[
   du = \frac{1}{1+x} \, dx \quad \text{and} \quad v = \arctan(x)
   \]

3. **Apply the integration by parts formula**:
   The integration by parts formula is:
   \[
   \int u \, dv = uv - \int v \, du
   \]
   So, applying it here:
   \[
   I = \left[\ln(1+x) \cdot \arctan(x)\right]_{0}^{1} - \int_{0}^{1} \arctan(x) \cdot \frac{1}{1+x} \, dx
   \]

4. **Evaluate the boundary term**:
   \[
   \left[\ln(1+x) \cdot \arctan(x)\right]_{0}^{1} = \left[\ln(2) \cdot \arctan(1)\right] - \left[\ln(1) \cdot \arctan(0)\right]
   \]
   Since \(\arctan(1) = \frac{\pi}{4}\) and \(\ln(1) = 0\):
   \[
   \left[\ln(1+x) \cdot \arctan(x)\right]_{0}^{1} = \ln(2) \cdot \frac{\pi}{4}
   \]

5. **Simplify the remaining integral**:
   We need to evaluate:
   \[
   \int_{0}^{1} \frac{\arctan(x)}{1+x} \, dx
   \]
   This integral is more challenging to evaluate directly, but it's known that:
   \[
   \int_{0}^{1} \frac{\arctan(x)}{1+x} \, dx = \frac{\pi}{8} \ln(2)
   \]

6. **Combine the results**:
   Thus, the original integral \( I \) is:
   \[
   I = \ln(2) \cdot \frac{\pi}{4} - \frac{\pi}{8} \ln(2)
   \]
   Simplifying, we get:
   \[
   I = \frac{\pi}{4} \ln(2) - \frac{\pi}{8} \ln(2) = \frac{\pi}{8} \ln(2)
   \]

So, the final answer is:
\[
\int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \, dx = \frac{\pi}{8} \ln(2)
\]

Would you like more details on any part of this solution, or do you have other questions?

---

**Related Questions:**
1. How can you derive the integral \(\int_{0}^{1} \frac{\arctan(x)}{1+x} \, dx\)?
2. What are some examples of integrals where integration by parts is particularly useful?
3. How does the arctangent function arise in integrals, and what are its key properties?
4. What are some advanced techniques for evaluating challenging integrals?
5. How can definite integrals be used in applications like calculating areas or solving physics problems?

**Tip:** When integration by parts results in a complex integral, it might help to recognize if the remaining integral is a known result or if further simplifications can be applied.

Learn how to solve the integral ∫(ln(1+x))/(1+x^2) from 0 to 1 using integration by parts. This problem involves integrating logarithmic and arctangent functions over a specific interval. Suitable for advanced learners and those interested in calculus problems.

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