To integrate $\frac{1}{1+x^2}$ from 0 to 1, we need to evaluate the definite integral:

$\int_{0}^{1} \frac{1}{1+x^2} \, dx$

The antiderivative of $\frac{1}{1+x^2}$ is the arctangent function, $\arctan(x)$. Therefore, the integral can be written as:

$\left. \arctan(x) \right|_0^1 = \arctan(1) - \arctan(0)$

Now, we evaluate the arctangent at the bounds:

$\arctan(1) = \frac{\pi}{4}, \quad \arctan(0) = 0$

So, the result of the integral is:

$\frac{\pi}{4} - 0 = \frac{\pi}{4}$

Thus, the value of the integral is $\boxed{\frac{\pi}{4}}$.

Would you like a more detailed explanation or have any questions?

Here are some related questions:

1. What is the derivative of $\arctan(x)$?
2. How do you find the integral of $\frac{1}{1-x^2}$?
3. Can you evaluate $\int_{0}^{\frac{\pi}{4}} \tan(x) \, dx$?
4. What is the relationship between arctan and the unit circle?
5. How do you compute the integral of $\frac{1}{x^2 + 4}$?

Tip: Understanding the basic antiderivatives, like that of $\frac{1}{1+x^2}$, helps in quickly solving integrals in calculus.