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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.

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  • $\begingroup$ WolframAlfa not working ? $\endgroup$ – Mariusz Iwaniuk Apr 20 '18 at 20:18
  • $\begingroup$ yes it does, but the answer is not simple $\endgroup$ – Zacky Apr 20 '18 at 20:22
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}

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  • $\begingroup$ nice sir! Can you tell me how did you thought of that substitution? Or is just experience ? $\endgroup$ – Zacky Apr 21 '18 at 17:22
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    $\begingroup$ @Zacky That's is an $\underline{\texttt{Euler Substitution}}$ . I just add the link at my answer top. Thanks. $\endgroup$ – Felix Marin Apr 21 '18 at 19:49

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