Compute $\int_{-1}^{1}\frac{a^{2x}+1}{a^{x}+b^{x}}dx$, where $a,b\in (1,\infty )$ I only found the value for $a=b$
2026-04-24 06:48:54.1777013334
Compute $\int_{-1}^{1}\frac{a^{2x}+1}{a^{x}+b^{x}}dx$
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Note $$I:=\int_{-1}^{1}\frac{a^{2x}+1}{a^{x}+b^{x}}dx=\int_{-1}^{1}\frac{a^{x}+a^{-x}}{1+(\frac{b}{a})^{x}}dx.\tag{1}$$ Under $x\to-x$, one has $$I=\int_{-1}^{1}\frac{a^{x}+a^{-x}}{1+(\frac{b}{a})^{-x}}dx=\int_{-1}^{1}\frac{(\frac{b}{a})^{x}(a^{x}+a^{-x})}{1+(\frac{b}{a})^{x}}dx.\tag{2}$$ Adding (1) to (2), one has $$ 2I=\int_{-1}^1(a^{x}+a^{-x})dx$$ which is easy to handle. You can do it.