Step-by-Step Solution
**Step 1: Simplify the integral**
Let $I = \int_{1}^{e}\frac{1}{\sqrt{4x^{2}-(x\log x)^{2}}}dx = \int_{1}^{e}\frac{1}{\sqrt{x^2(4-(\log x)^{2})}}dx = \int_{1}^{e}\frac{1}{x\sqrt{4-(\log x)^{2}}}dx$
**Step 2: Apply substitution**
Let $u = \log x$. Then, $\frac{du}{dx} = \frac{1}{x}$, so $du = \frac{1}{x}dx$.
When $x = 1$, $u = \log 1 = 0$.
When $x = e$, $u = \log e = 1$.
Therefore, $I = \int_{0}^{1}\frac{1}{\sqrt{4-u^{2}}}du$
**Step 3: Evaluate the integral**
We know that $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}(\frac{x}{a}) + C$.
So, $I = \int_{0}^{1}\frac{1}{\sqrt{2^{2}-u^{2}}}du = \left[\sin^{-1}\left(\frac{u}{2}\right)\right]_{0}^{1} = \sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}(0)$
**Step 4: Find the final value**
Since $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ and $\sin^{-1}(0) = 0$,
$I = \frac{\pi}{6} - 0 = \frac{\pi}{6}$