How to evaluate the integral $\int_0^2 \frac{\ln x}{\sqrt {x^2-2x+2}}dx$?

Question

Yesterday's integral may be too difficult, I think the following integral should not be difficult. $$I=\int_0^2 \frac{\ln x}{\sqrt {x^2-2x+2}}dx=\int_{-1}^1\frac{\ln(x+1)}{\sqrt {x^2+1}}dx=\int_{-1}^1\ln(x+1)d(\ln (x+\sqrt {x^2+1}))$$ $$=\ln (x+1)\ln (x+\sqrt {x^2+1})|_{-1}^1-\int_{-1}^1\frac{\ln (x+\sqrt {x^2+1})}{(x+1)}dx=......$$ So far, I don't know what to do, any help is appreciated.

Answer 1

Let $\sqrt {{x^2} - 2x + 2} = x + t$, then we have $$I=\int_0^2 {\frac{{\ln x}}{{\sqrt {{x^2} - 2x + 2} }}} = \int_{\sqrt 2 - 2}^{\sqrt 2 } {\ln \left( {\frac{{2 - {t^2}}}{{2(t + 1)}}} \right)\frac{1}{{1 + t}}dt} $$ This integral can be broken down into pieces of the form $\int \frac{\ln x}{x+a} dx$, which can be evaluated by dilogarithm. The final result being $$\DeclareMathOperator{\Li}{Li} I=\frac{{{\pi ^2}}}{{12}} - \frac{9\ln ^22}{8} + \frac{3}{2}\ln (1 + \sqrt 2 )\ln 2 - \frac{1}{2}{\ln ^2}(1 + \sqrt 2 ) - \Li_2\left( {2\sqrt 2 - 2} \right) - \Li_2\left( {\frac{{2 - \sqrt 2 }}{4}} \right)$$