Answer: \(\displaystyle \int_{6}^{7} \frac{1}{\sqrt{(x-5)^2-1}}\,dx=\ln(2+\sqrt{3})\).

So \(a=2\) and \(b=3\).

Use the standard result \(\displaystyle \int \frac{1}{\sqrt{u^2-1}}\,du=\cosh^{-1}(u)+C\).

Let \(u=x-5\). Then \(du=dx\), so

\(\displaystyle \int_{6}^{7} \frac{1}{\sqrt{(x-5)^2-1}}\,dx=\left[\cosh^{-1}(x-5)\right]_{6}^{7}.\)

Now apply the limits:

\(\displaystyle =\cosh^{-1}(2)-\cosh^{-1}(1).\)

Since \(\cosh^{-1}(1)=0\), this is

\(\displaystyle \cosh^{-1}(2).\)

Using \(\cosh^{-1}(y)=\ln\big(y+\sqrt{y^2-1}\big)\),

\(\displaystyle \cosh^{-1}(2)=\ln\big(2+\sqrt{4-1}\big)=\ln(2+\sqrt{3}).\)

Therefore, \(\displaystyle \int_{6}^{7} \frac{1}{\sqrt{(x-5)^2-1}}\,dx=\ln(2+\sqrt{3})\).