Let $I=\int \frac{10^{\frac{x}{2}}}{10^{-x}-10^x} d x$
$$
\begin{aligned}
& =\int \frac{10^{\frac{x}{2}}}{\sqrt{\frac{1}{10^x}-10^x}} d x=\int \frac{10^{\frac{x}{2}}}{\sqrt{\frac{1-\left(10^x\right)^2}{10^x}}} d x \\
& =\int \frac{10^{\frac{x}{2}} \cdot 10^{\frac{x}{2}}}{\sqrt{1-\left(10^x\right)^2}} d x=\int \frac{10^x}{\sqrt{1-\left(10^x\right)^2}} d x
\end{aligned}
$$
Put $10^{\mathrm{x}}=\mathrm{t} \Rightarrow 10^{\mathrm{x}}(\log 10) \mathrm{dx}=\mathrm{dt}$
$$
\therefore \mathrm{I}=\frac{1}{\log 10} \int \frac{\mathrm{dt}}{\sqrt{1-\mathrm{t}^2}}=\frac{1}{\log 10} \sin ^{-1}(\mathrm{t})+\mathrm{c}=\frac{1}{\log 10} \sin ^{-1}\left(10^{\mathrm{x}}\right)+\mathrm{c}
$$