An integral related with the Riemann $\zeta$ function

Reverse sum and integral and show that

$$\int_0^{\infty} \frac{dx}{a^x+b} = \frac{\log{(1+b)}}{b \log{a}} $$

Note that $a=1+k^s$ and $b=k^s$. The logs cancel and you are just left with the sum over $k^{-s}$, which is the stated result.

Just to complete Ron Gordon's excellent answer, notice that: $$ I(a,b)=\int_{0}^{+\infty}\frac{dx}{e^{x\log a}+b} = \frac{1}{\log a}\int_{0}^{+\infty}\frac{dz}{e^z+b}=\frac{1}{\log a}\int_{1}^{+\infty}\frac{dt}{t(t+b)}$$ and: $$ \int_{1}^{+\infty}\frac{dt}{t(t+b)}=\int_{0}^{1}\frac{du}{1+bu}=\frac{1}{b}\int_{0}^{b}\frac{dv}{1+v}=\frac{\log(1+b)}{b}$$ through the substitutions $x=\frac{z}{\log a}$, $z=\log t$, $t=\frac{1}{u}$ and $u=\frac{v}{b}$. That leads to: $$\int_{0}^{+\infty}\sum_{k\geq 1}\frac{dx}{(k^s+1)^x+k^s}=\sum_{k\geq 1}\frac{\log(1+k^s)}{k^s \log(1+k^s)} = \zeta(s)$$ as wanted.