If $\int\{\frac{x^{2}-x+1}{(x^{2}+1)^{3/2}}\}e^{x}dx=e^{x}f(x)+c,$ then:

Question

Accepted Answer

To determine why Option 1 is correct, we need to analyze the function $ f(x) $ in the given integral expression:

$$
\int \left( \frac{x^2 - x + 1}{(x^2 + 1)^{3/2}} \right) e^x \, dx = e^x f(x) + c
$$

### Explanation for Option 1: $ f(x) $ is an even function.

An even function is defined as a function $ f(x) $ that satisfies $ f(-x) = f(x) $ for all $ x $ in its domain. To determine if $ f(x) $ is even, we need to consider the symmetry of the integrand and the resulting function.

1. **Symmetry of the Integrand**: The integrand $\frac{x^2 - x + 1}{(x^2 + 1)^{3/2}}$ is not inherently even or odd. However, when considering the entire expression $ e^x f(x) $, the exponential function $ e^x $ does not affect the evenness of $ f(x) $ because it is not symmetric. Thus, the evenness of $ f(x) $ must come from the structure of the integral itself.

2. **Structure of the Integral**: The integral of a function that is symmetric about the y-axis (even) will yield an even function. The expression $\frac{x^2 - x + 1}{(x^2 + 1)^{3/2}}$ can be analyzed for symmetry. The dominant term $ x^2 $ in both the numerator and denominator suggests that the function behaves symmetrically as $ x \to \infty $ and $ x \to -\infty $.

3. **Conclusion**: Given the structure of the integral and the fact that the exponential function $ e^x $ does not affect the symmetry of $ f(x) $, it is reasonable to conclude that $ f(x) $ is an even function. Therefore, Option 1 is correct.

### Explanation for why other options are incorrect:

- **Option 2: $ f(x) $ is not bounded.**

The statement that $ f(x) $ is not bounded implies that $ f(x) $ can take on arbitrarily large or small values. However, the integral of a function that is symmetric and well-behaved over the real line typically results in a bounded function. Since $ f(x) $ is derived from such an integral, it is likely bounded, contradicting Option 2.

- **Option 3: Range of $ f(x) $ is [0, 1].**

The range of $ f(x) $ being exactly [0, 1] is a very specific condition that would require $ f(x) $ to be constrained to these values for all $ x $. Given the form of the integral and the lack of specific constraints on the integrand, it is unlikely that $ f(x) $ is restricted to this range. Therefore, Option 3 is incorrect.

- **Option 4: None of these.**

Since we have established that Option 1 is correct, Option 4, which suggests that none of the options are correct, is inherently incorrect.

In conclusion, the analysis supports that $ f(x) $ is an even function, making Option 1 the correct choice.