Integrate $\int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{\ln x}{x\sqrt{-x^2+6x-1}}$ How to integrate $$\int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{\ln x}{x\sqrt{-x^2+6x-1}}\, dx$$

I believe the exact value of this integral cannot be found , however there could be a way I am not aware of. If not, is it possible to put into a familiar form using other functions such as $Li(x)$ for example. 

After some attempts i am still lost and Mathematica is not too helpful.

Thank you kindly for your help and time.
Substitute $x=\frac1{3-2\sqrt2\cos t}$ to get

\begin{align}
&amp; \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{\ln x}{x\sqrt{-x^2+6x-1}}\, dx\\
= &amp; \int_0^\pi \ln(3-2\sqrt2 \cos t) dt\\
=&amp; \int_0^\pi [\ln2+\ln(1+  \frac12-\frac2{\sqrt2 }\cos t) ]dt=\pi\ln2
\end{align}
where $\int_0^\pi \ln(1+ a^2-2a\cos t)dt=0$ with $a=\frac1{\sqrt2}$ is used.
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
hint

Put $$x=\frac 1t$$

and observe that
$$\frac{1}{3-2\sqrt{2}}=3+2\sqrt{2}$$

The integral becomes
$$\int_{3-\sqrt{8}}^{3+\sqrt{8}}\frac{\ln(t)}{\sqrt{8-(t-3)^2}}dt$$
Now, we put
$$(t-3)=\sqrt{8}\sin(u)$$
with
$$dt=\sqrt{8}\cos(u)du$$

this yields to

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(3+\sqrt{8}\sin(u))du$$