We are given $f'(x) = \frac{1}{3 - x^2}$ and $f(0) = 1$. We need to find an interval in which $f(1)$ lies.

To solve this, we first integrate $f'(x)$ to find $f(x)$.

Step 1: Integrate $f'(x)$

We integrate $f'(x) = \frac{1}{3 - x^2}$. Using partial fraction decomposition:

$\frac{1}{3 - x^2} = \frac{1}{( \sqrt{3} - x)(\sqrt{3} + x)}$

However, it's easier to recognize that this integral can be directly evaluated as a standard form:

$\int \frac{1}{3 - x^2} \, dx = \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + x}{\sqrt{3} - x} \right| + C$

Step 2: Solve for $C$ using the initial condition

We know $f(0) = 1$. Substituting into the equation we get:

$f(0) = \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + 0}{\sqrt{3} - 0} \right| + C = 1$

This simplifies to:

$1 = \frac{1}{2\sqrt{3}} \ln(1) + C \implies C = 1$

Thus, the function becomes:

$f(x) = \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + x}{\sqrt{3} - x} \right| + 1$

Step 3: Calculate $f(1)$

Now, substitute $x = 1$ into the equation for $f(x)$:

$f(1) = \frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right| + 1$

Let’s compute this logarithm and simplify:

$f(1) = \frac{1}{2\sqrt{3}} \ln \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right) + 1$

The value of $f(1)$ lies within a specific range, but based on this calculation, $f(1)$ is slightly greater than 1.5.

Step 4: Conclusion

Therefore, the value of $f(1)$ lies in the interval $(1, 2)$.

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Would you like more details on the logarithmic simplification or the exact value of $f(1)$?

Here are some related questions:

1. How can we evaluate $f'(x)$ at specific points?
2. What is the domain of $f(x)$ based on the derivative?
3. How does the behavior of $f(x)$ change near $x = \pm \sqrt{3}$?
4. Can we use a numerical method to approximate the exact value of $f(1)$?
5. What are partial fraction decomposition techniques for more complex rational functions?

Tip: When solving integrals involving logarithms, always double-check the domain to avoid introducing undefined values due to the logarithmic expression.