Integral $\int_{\frac{1}{2}}^1 \frac{\ln x}{\sqrt{x^2+1}}dx$ I am trying to evaluate $$I=\int_{1/2}^1 \frac{\log(x)}{\sqrt{x^2+1}}\,dx$$ My attempt was to integrate by parts with $u=\log(x)$ and $v=\log{(x+\sqrt{1+x^2}})$ to get$$I=-\log{\frac{1}{2}}\log{\left(\frac{1}{2}+\sqrt{1+\frac{1}{4}}\right)}-\int_{1/2}^1 \frac{\log{(x+\sqrt{1+x^2}})}{x}\,dx$$ so simplifying the integral is $$I=\log{(2\phi)}-I_1$$ where $I_1=\int_{\frac{1}{2}}^1 \frac{\sinh^{-1}(x)}{x}\,dx$ If possible I would love to get some help in evaluating those.
I also thought about considering $I(k)=\int_{\frac{1}{2}}^1 \frac{x^k} {\sqrt{x^2+1}}\,dx$ writting the denominator as a binomial series, integrate it then derivate and plug $k=0$ to get the answer but the series is not nice.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With Euler Substitution

$\ds{x \equiv {1 - t^{2} \over 2t} \implies t = \root{x^{2} + 1} - x\quad}$ and
  $\ds{\quad \pars{~a \equiv {\root{5} - 1 \over 2}\,,\
b \equiv \root{2} - 1~}}$:


\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{1/2}^{1}{\ln\pars{x} \over \root{x^{2} + 1}}\,\dd x}} =
-\int_{a}^{b}\ln\pars{1 - t^{2} \over 2t}\,{\dd t \over t}
=
-\int_{a}^{b}{\ln\pars{1 - t^{2}} \over t}\,\dd t +
\int_{a}^{b}{\ln\pars{2t} \over t}\,\dd t
\\[5mm] = &\
-\,{1 \over 2}\int_{a^{2}}^{b^{2}}{\ln\pars{1 - t} \over t}\,\dd t +
\int_{2a}^{2b}{\ln\pars{t} \over t}\,\dd t
\\[5mm] & =
{1 \over 2}\,\mrm{Li}_{2}\pars{b^{2}} - {1 \over 2}\,\mrm{Li}_{2}\pars{a^{2}} +
{1 \over 2}\,\ln^{2}\pars{2b} - {1 \over 2}\,\ln^{2}\pars{2a}
\\[5mm] & = \bbx{%
{1 \over 2}\,\mrm{Li}_{2}\pars{3 - 2\root{2}} -
{1 \over 2}\,\mrm{Li}_{2}\pars{3 - \root{5} \over 2} +
{1 \over 2}\,\ln^{2}\pars{2\root{2} - 2} -
{1 \over 2}\,\ln^{2}\pars{\root{5} - 1}}
\\[5mm] \approx &\ -0.1282
\end{align}
