40 The set of integer solutions to the inequality $\log _{6}(1+\sqrt{x})>\log _{25} x$ is
$40\{x \mid 0<t$. That is, $1+5^{t}>6^{t},\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}>1$. Since $f(t)=\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}$ is a decreasing function on the real number set $R$, and $f(1)=1$, it follows from $f(t)>f(1)$ that $t<1$, i.e., $\log _{25} x<1,0<x<25$. The solution set is $\{x \mid 0<x<25, x \in \mathbf{Z}\}$.