Let's evaluate the integral:

$\int_0^1 \ln(1 + x^2) \, dx$

We will use integration by parts to solve this integral. The formula for integration by parts is:

$\int u \, dv = uv - \int v \, du$

First, we need to choose $u$ and $dv$. For this integral, let:

$u = \ln(1 + x^2) \quad \text{and} \quad dv = dx$

Then, we need to compute $du$ and $v$:

$du = \frac{d}{dx}[\ln(1 + x^2)] \, dx = \frac{2x}{1 + x^2} \, dx$

$v = \int dv = \int dx = x$

Now, apply the integration by parts formula:

$\int_0^1 \ln(1 + x^2) \, dx = \left. x \ln(1 + x^2) \right|_0^1 - \int_0^1 x \cdot \frac{2x}{1 + x^2} \, dx$

Evaluate the boundary term:

$\left. x \ln(1 + x^2) \right|_0^1 = (1 \cdot \ln(1 + 1^2)) - (0 \cdot \ln(1 + 0^2)) = \ln(2) - 0 = \ln(2)$

Now, simplify the remaining integral:

$\int_0^1 \frac{2x^2}{1 + x^2} \, dx$

Rewrite the integrand:

$\frac{2x^2}{1 + x^2} = 2 - \frac{2}{1 + x^2}$

Thus, the integral becomes:

$\int_0^1 \left(2 - \frac{2}{1 + x^2}\right) \, dx = \int_0^1 2 \, dx - 2 \int_0^1 \frac{1}{1 + x^2} \, dx$

Evaluate each integral separately:

$\int_0^1 2 \, dx = 2x \bigg|_0^1 = 2(1) - 2(0) = 2$

$\int_0^1 \frac{1}{1 + x^2} \, dx = \left. \arctan(x) \right|_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$

So the integral is:

$2 - 2 \cdot \frac{\pi}{4} = 2 - \frac{\pi}{2}$

Combine everything together:

$\int_0^1 \ln(1 + x^2) \, dx = \ln(2) - \left(2 - \frac{\pi}{2}\right) = \ln(2) - 2 + \frac{\pi}{2}$

Thus, the value of the integral is:

$\boxed{\ln(2) - 2 + \frac{\pi}{2}}$

Would you like to see more details or have any questions about this solution?

Here are 8 related questions to expand your understanding:

1. How can the method of integration by parts be applied to other types of integrals?
2. What are some common techniques for integrating logarithmic functions?
3. Can you evaluate $\int \ln(1 + x^2) \, dx$ without specified limits?
4. What are some other definite integrals involving logarithmic functions that can be solved using integration by parts?
5. How does the substitution method help in solving integrals involving logarithms?
6. Can you derive other properties of the natural logarithm function that can be useful in integration?
7. How do integrals involving inverse trigonometric functions typically get evaluated?
8. What are some real-world applications of integrals involving logarithmic functions?

Tip: When solving integrals, always look for opportunities to simplify the problem using algebraic manipulations, substitutions, or known integral formulas.