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} & \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{\ln x}{x\sqrt{-x^2+6x-1}}\, dx\\ = & \int_0^\pi \ln(3-2\sqrt2 \cos t) dt\\ =& \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 $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
Letting $x\mapsto \frac{1}{x} $ transforms the integral $$ \begin{aligned} I & =-\int_\alpha^\beta \frac{\ln x}{\sqrt{-x^2+6 x-1}} d x \\ & =-\int_\alpha^\beta \frac{\ln x}{\sqrt{8-(x-3)^2}} d x, \end{aligned} $$ where $\alpha, \beta = 3\mp 2\sqrt 2$ and $\alpha \beta =1.$
Putting $x-3=\sqrt{8} \cos \theta$ changes $$ I= \int_0^\pi \ln (3+\sqrt{8} \cos \theta) d \theta $$ By my post, we have $$ \boxed{I =\pi \ln \left(\frac{3+\sqrt{3^2-(\sqrt{8})^2}}{2}\right) =\pi \ln 2} $$
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
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(3+\sqrt{8}\sin(u))du$$
\ln. For operators that don't have a command of their own, you can use\operatorname{name}. – joriki May 12 '20 at 23:31