微积分(A)随缘一题[33]

计算：\(\int_{-1}^{1}\frac{x^{2020}+1}{2020^x+1}dx\)

考虑：\(\int_{-a}^{a} f(x)dx=\int_0^{a}f(x)dx+\int_{-a}^{0}f(x)dx=\int_0^a[f(x)+f(-x)]dx\)

\[\int_{-1}^{1}\frac{x^{2020}+1}{2020^x+1}dx=\int_0^1 \left[\frac{x^{2020}+1}{2020^x+1}+\frac{x^{2020}+1}{2020^{-x}+1}\right]dx=\int_0^1 [x^{2020}+1]dx=\left(\frac{x^{2021}}{2021}+x\right) \bigg|_0^1=\frac{2022}{2021} \]