Question

$\int_{0}^{1} \frac{x^{2023}+x^{2025}}{(1+x+x^2)(1-x+x^2)-1}dx$

Answer 1

1/2022

Answer 2

The problem asks to evaluate the definite integral $\int_{0}^{1} \frac{x^{2023}+x^{2025}}{(1+x+x^2)(1-x+x^2)-1}dx$.

First, simplify the denominator: The expression $(1+x+x^2)(1-x+x^2)$ can be written in the form $(A+B)(A-B)$, where $A = (1+x^2)$ and $B = x$. So, $(1+x+x^2)(1-x+x^2) = ((1+x^2)+x)((1+x^2)-x)$ $= (1+x^2)^2 - x^2$ $= (1+2x^2+x^4) - x^2$ $= 1+x^2+x^4$

Now, substitute this back into the denominator expression: $(1+x+x^2)(1-x+x^2)-1 = (1+x^2+x^4)-1$ $= x^2+x^4$ $= x^2(1+x^2)$

Next, simplify the numerator: $x^{2023}+x^{2025} = x^{2023}(1+x^2)$

Now, substitute the simplified numerator and denominator back into the integral: The integrand becomes $\frac{x^{2023}(1+x^2)}{x^2(1+x^2)}$. For $x \in [0,1]$, the term $(1+x^2)$ is never zero, so it can be cancelled from the numerator and denominator. Also, for $x \in (0,1]$, $x^2$ is not zero, so $\frac{x^{2023}}{x^2} = x^{2021}$. Even though the original integrand is of the form $\frac{0}{0}$ at $x=0$, the simplified integrand $x^{2021}$ is continuous over $[0,1]$. Thus, the integral can be evaluated using the simplified form.

So the integral simplifies to: $\int_{0}^{1} x^{2021}dx$

Now, apply the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$: $\int_{0}^{1} x^{2021}dx = \left[\frac{x^{2021+1}}{2021+1}\right]_{0}^{1}$ $= \left[\frac{x^{2022}}{2022}\right]_{0}^{1}$

Finally, apply the limits of integration: $= \frac{1^{2022}}{2022} - \frac{0^{2022}}{2022}$ $= \frac{1}{2022} - 0$ $= \frac{1}{2022}$

The final answer is $\frac{1}{2022}$.

Explanation of the solution:

1. Simplify the denominator $(1+x+x^2)(1-x+x^2)-1$ to $x^2(1+x^2)$ by recognizing the difference of squares pattern and simplifying.
2. Factor the numerator $x^{2023}+x^{2025}$ as $x^{2023}(1+x^2)$.
3. Cancel the common term $(1+x^2)$ and simplify the power of $x$ in the integrand, resulting in $\int_{0}^{1} x^{2021}dx$.
4. Integrate $x^{2021}$ using the power rule to get $\frac{x^{2022}}{2022}$.
5. Evaluate the definite integral by applying the limits from $0$ to $1$, which yields $\frac{1}{2022}$.