Answer
$$\frac{1}{\ln 5}\sec^{-1}5^x+c$$
Work Step by Step
Let $u=5^x$, then $du=5^x\ln 5 dx$, now we have
$$\int \frac{dx}{\sqrt{5^{2x}-1}}=\frac{1}{\ln 5}\int \frac{du}{u\sqrt{u^2-1}} \\
=\frac{1}{\ln 5}\sec^{-1}u+c=\frac{1}{\ln 5}\sec^{-1}5^x+c.$$
