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Question: $\int \frac{x^{2023}+x^{2025}}{(1+x+x^2)(1-x+x^2)-1}dx$...

∫x2023+x2025(1+x+x2)(1−x+x2)−1dx\int \frac{x^{2023}+x^{2025}}{(1+x+x^2)(1-x+x^2)-1}dx

Answer

x20222022+C\frac{x^{2022}}{2022} + C

Explanation

Solution

Explanation of the solution:

  1. Simplify the Denominator:
    The denominator is (1+x+x2)(1−x+x2)−1(1+x+x^2)(1-x+x^2)-1.
    Recognize the pattern (A+B)(A−B)=A2−B2(A+B)(A-B) = A^2-B^2 by letting A=(1+x2)A = (1+x^2) and B=xB = x.
    So, (1+x+x2)(1−x+x2)=((1+x2)+x)((1+x2)−x)(1+x+x^2)(1-x+x^2) = ((1+x^2)+x)((1+x^2)-x)
    =(1+x2)2−x2= (1+x^2)^2 - x^2
    =(1+2x2+x4)−x2= (1+2x^2+x^4) - x^2
    =1+x2+x4= 1+x^2+x^4
    Now substitute this back into the denominator expression:
    (1+x2+x4)−1=x2+x4=x2(1+x2)(1+x^2+x^4)-1 = x^2+x^4 = x^2(1+x^2).

  2. Simplify the Numerator:
    The numerator is x2023+x2025x^{2023}+x^{2025}.
    Factor out the common term x2023x^{2023}:
    x2023+x2025=x2023(1+x2)x^{2023}+x^{2025} = x^{2023}(1+x^2).

  3. Simplify the Integrand:
    Substitute the simplified numerator and denominator back into the integral:
    The integrand becomes x2023(1+x2)x2(1+x2)\frac{x^{2023}(1+x^2)}{x^2(1+x^2)}.
    For x≠0x \ne 0, the term (1+x2)(1+x^2) can be cancelled from the numerator and denominator (since 1+x21+x^2 is never zero for real xx).
    Also, x2023x2=x2023−2=x2021\frac{x^{2023}}{x^2} = x^{2023-2} = x^{2021}.
    Thus, the integral simplifies to ∫x2021dx\int x^{2021}dx.

  4. Evaluate the Indefinite Integral:
    Apply the power rule for integration, ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C:
    ∫x2021dx=x2021+12021+1+C\int x^{2021}dx = \frac{x^{2021+1}}{2021+1} + C
    =x20222022+C= \frac{x^{2022}}{2022} + C.