To solve the problem, let's carefully analyze the mathematical expression:

We are given: $f(x) = \frac{1}{1+x^2} + \sqrt{1-x^2} \int_0^1 f(x) \, dx$

Our goal is to find the value of $\int_0^1 f(x) \, dx$.

Step-by-Step Solution:

1. Set the unknown integral:
   Let: $I = \int_0^1 f(x) \, dx$ This makes $f(x)$ expressed as: $f(x) = \frac{1}{1+x^2} + \sqrt{1-x^2} \cdot I$

2. Integrate $f(x)$ over $[0, 1]$:
   Substitute $f(x)$ into the integral: $I = \int_0^1 \left( \frac{1}{1+x^2} + \sqrt{1-x^2} \cdot I \right) dx$

   Separate the terms: $I = \int_0^1 \frac{1}{1+x^2} \, dx + \int_0^1 \sqrt{1-x^2} \cdot I \, dx$

   The first term, $\int_0^1 \frac{1}{1+x^2} \, dx$, is a standard integral: $\int_0^1 \frac{1}{1+x^2} \, dx = \arctan(1) - \arctan(0) = \frac{\pi}{4}.$

   For the second term: $\int_0^1 \sqrt{1-x^2} \cdot I \, dx = I \cdot \int_0^1 \sqrt{1-x^2} \, dx$ The integral $\int_0^1 \sqrt{1-x^2} \, dx$ represents the area of a quarter-circle with radius $1$, which is: $\int_0^1 \sqrt{1-x^2} \, dx = \frac{\pi}{4}.$

   Substitute this back: $I = \frac{\pi}{4} + I \cdot \frac{\pi}{4}.$

3. Solve for $I$:
   Factor out $I$ on the right-hand side: $I = \frac{\pi}{4} + \frac{\pi}{4} \cdot I$ Rearrange: $I \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4}.$ Simplify: $I = \frac{\frac{\pi}{4}}{1 - \frac{\pi}{4}}.$ Combine into a single fraction: $I = \frac{\pi}{4 - \pi}.$

Final Answer:

The correct choice is: $\boxed{B. \frac{\pi}{4-\pi}}$

Would you like further clarification on this solution? Here are some related questions to explore:

1. How do we interpret $\int_0^1 \sqrt{1-x^2} \, dx$ as the area of a geometric shape?
2. What are some other examples of standard integrals involving $\arctan(x)$?
3. Can this problem be solved numerically instead of analytically? Why or why not?
4. How would the solution change if the limits of integration were different?
5. Why does the appearance of $I$ on both sides lead to a solvable equation?

Tip: Memorizing standard integrals like $\int \frac{1}{1+x^2} dx$ and $\int \sqrt{1-x^2} dx$ is crucial for solving problems efficiently!