Evaluate the given expression.

$$\int \frac{\sqrt{81-(\log x)^2}}{x} d x$$​​

Correct option is C

​$$\begin{aligned}&\text{Let } \log x = t \Rightarrow \frac{d}{dx}(\log x) = \frac{1}{x} \Rightarrow dx = x \, dt. \\[6pt]&\text{Then the given integral becomes:} \\&\int \frac{\sqrt{81 - (\log x)^2}}{x} \, dx = \int \sqrt{81 - t^2} \, dt \\[10pt]&\text{We use the standard formula:} \\&\int \sqrt{a^2 - t^2} \, dt = \frac{t}{2} \sqrt{a^2 - t^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{t}{a} \right) + C \\[10pt]&\text{Here } a = 9, \text{ so we get:} \\&\int \sqrt{81 - t^2} \, dt = \frac{t}{2} \sqrt{81 - t^2} + \frac{81}{2} \sin^{-1} \left( \frac{t}{9} \right) + C \\[10pt]&\text{Substitute back } t = \log x,\ \text{we get the final answer:} \\[6pt]&{\frac{\log x}{2} \sqrt{81 - (\log x)^2} + \frac{81}{2} \sin^{-1} \left( \frac{\log x}{9} \right) + C}\end{aligned}$$​​