1. **State the problem:** We want to find the integral of the function $\sin^{-1}(\log x)$, which means we want to compute $$\int \sin^{-1}(\log x) \, dx.$$\n\n2. **Recall the formula and rules:** The integral involves the inverse sine function composed with the logarithm function. We will use integration by parts, where \(u = \sin^{-1}(\log x)\) and \(dv = dx\).\n\n3. **Set up integration by parts:**\nLet \(u = \sin^{-1}(\log x)\) so that \(du = \frac{1}{\sqrt{1-(\log x)^2}} \cdot \frac{1}{x} dx\) by the chain rule.\nLet \(dv = dx\) so that \(v = x\).\n\n4. **Apply integration by parts formula:**\n$$\int u \, dv = uv - \int v \, du$$\nSo,\n$$\int \sin^{-1}(\log x) \, dx = x \sin^{-1}(\log x) - \int x \cdot \frac{1}{\sqrt{1-(\log x)^2}} \cdot \frac{1}{x} dx = x \sin^{-1}(\log x) - \int \frac{1}{\sqrt{1-(\log x)^2}} dx.$$\n\n5. **Simplify the remaining integral:**\nWe now need to evaluate $$\int \frac{1}{\sqrt{1-(\log x)^2}} dx.$$\nUse substitution \(t = \log x\), so \(dt = \frac{1}{x} dx \Rightarrow dx = x dt = e^t dt\).\n\n6. **Rewrite the integral in terms of \(t\):**\n$$\int \frac{1}{\sqrt{1-t^2}} e^t dt.$$\n\n7. **Final integral:**\nThe integral becomes $$\int \frac{e^t}{\sqrt{1-t^2}} dt,$$ which does not have a simple elementary antiderivative.\n\n**Therefore, the integral is:**\n$$\int \sin^{-1}(\log x) \, dx = x \sin^{-1}(\log x) - \int \frac{e^t}{\sqrt{1-t^2}} dt + C,$$ where \(t = \log x\) and \(C\) is the constant of integration.\n\nThis is the most simplified form expressible in elementary functions.