Question

If $f(x) = \int \frac{x^8+4}{x^4-2x^2+2} dx$ and $f(0) = 0$, then:

• A. $f(x)$ is an odd function
• B. $f(x)$ has range $R$
• C. $f(x)$ has at least one real root
• D. $f(x)$ is a monotonic function

Answer 1

Answer: (A), (B), (C), (D)

All of the above statements are correct.

Answer 2

1. Factorize the numerator: The numerator $x^8+4$ can be factorized as $(x^4-2x^2+2)(x^4+2x^2+2)$.

2. Simplify the integrand: The integrand becomes $\frac{(x^4-2x^2+2)(x^4+2x^2+2)}{x^4-2x^2+2}$. The denominator $x^4-2x^2+2 = (x^2-1)^2+1$ is always positive and non-zero. Thus, the simplified integrand is $x^4+2x^2+2$.

3. Integrate to find $f(x)$: $f(x) = \int (x^4+2x^2+2) dx = \frac{x^5}{5} + \frac{2x^3}{3} + 2x + C$.

4. Determine the constant of integration $C$: Given $f(0)=0$: $f(0) = \frac{0^5}{5} + \frac{2(0)^3}{3} + 2(0) + C = 0 \implies C = 0$. So, $f(x) = \frac{x^5}{5} + \frac{2x^3}{3} + 2x$.

5. Analyze the properties of $f(x)$:

   • Odd function: $f(-x) = \frac{(-x)^5}{5} + \frac{2(-x)^3}{3} + 2(-x) = -\frac{x^5}{5} - \frac{2x^3}{3} - 2x = -f(x)$. Thus, $f(x)$ is an odd function.
   • Range $R$: $f(x)$ is a polynomial of odd degree (5). As $x \to \infty$, $f(x) \to \infty$, and as $x \to -\infty$, $f(x) \to -\infty$. Being continuous, its range is $(-\infty, \infty)$ or $R$.
   • At least one real root: Since $f(0)=0$, $x=0$ is a real root.
   • Monotonic function: The derivative is $f'(x) = x^4+2x^2+2$. Since $x^4 \ge 0$ and $x^2 \ge 0$, $f'(x) = (x^2)^2 + 2(x^2) + 2 \ge 2$. As $f'(x) > 0$ for all real $x$, $f(x)$ is strictly increasing and hence monotonic.