Answer: \(\displaystyle \int_{2}^{\frac{7}{2}} \frac{1}{\sqrt{4x-x^2-1}}\,dx=\frac{\pi}{3}\).

Complete the square inside the square root:

\(4x-x^2-1=3-(x-2)^2\).

So

\(\displaystyle \int_{2}^{\frac{7}{2}} \frac{1}{\sqrt{4x-x^2-1}}\,dx=\int_{2}^{\frac{7}{2}} \frac{1}{\sqrt{3-(x-2)^2}}\,dx\).

Let \(u=x-2\). Then \(du=dx\), and the limits become \(u=0\) and \(u=\frac32\).

Hence the integral is

\(\displaystyle \int_{0}^{\frac32} \frac{1}{\sqrt{3-u^2}}\,du\).

Using \(\displaystyle \int \frac{1}{\sqrt{a^2-u^2}}\,du=\sin^{-1}\!\left(\frac{u}{a}\right)+C\), with \(a=\sqrt3\), this becomes

\(\displaystyle \left[\sin^{-1}\!\left(\frac{u}{\sqrt3}\right)\right]_{0}^{\frac32}\).

Therefore

\(\displaystyle \sin^{-1}\!\left(\frac{3}{2\sqrt3}\right)-\sin^{-1}(0)=\sin^{-1}\!\left(\frac{\sqrt3}{2}\right)=\frac{\pi}{3}\).

So the exact value is \(\boxed{\frac{\pi}{3}}\).