Q:

Compute $\int_{-1}^{1}\frac{a^{2x}+1}{a^{x}+b^{x}}dx$

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 $a=b$

A:

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.

