$$I=\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx$$

My attempt:$$I=\int_{0}^{1}\log x d(\log(x+\sqrt{1+x^2}))$$ $$=\log x\log(x+\sqrt{1+x^2})|_0^1-\int_{0}^{1}\frac{\log(x+\sqrt{1+x^2})}{x}dx$$ I don't know how to proceed below, please help me. That is different to me.

  • $\begingroup$ Do you think this possible to analytically integrate? If so, why? WolframAlpha seems to disagree... $\endgroup$
    – gt6989b
    Sep 3, 2018 at 3:45
  • 2
    $\begingroup$ @DarkKnight The denominator is $$\frac {\log x}{\color{red}{\sqrt{1+x^2}}}$$ though. $\endgroup$
    – Frank W
    Sep 3, 2018 at 3:49
  • 2
    $\begingroup$ @gt6989b Mathematica gives the closed form $$I = \text{Li}_2\left(-\sqrt{2}\right)-\text{Li}_2\left(-1-\sqrt{2}\right)-\frac{\pi ^2}{12}+\frac{1}{2} \log \left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right) \sinh ^{-1}(1)$$ $\endgroup$
    – user85362
    Sep 3, 2018 at 3:55
  • $\begingroup$ It has no analytical form. I suggest you use Taylor in $1/\sqrt{1+x^2}$, so you can get an approximation. $\endgroup$
    – Joda
    Sep 3, 2018 at 4:35

2 Answers 2


$\newcommand{\arcsinh}{\operatorname{arcsinh}}$This problem shows how powerful Euler Substitutions are and how quickly they can be used to decompose a difficult integral into something much more manageable.

First off, make the Euler Substitution $\sqrt{x^2+1}=-x+t$. Therefore, it's easy to see that$$\mathrm dx=\frac {t^2+1}{2t^2}\,\mathrm dt\qquad\qquad x=\frac {t^2-1}{2t}$$And thus, the integral becomes$$\mathfrak{I}=\int\limits_1^{1+\sqrt2}\mathrm dt\,\frac {\log(t^2-1)}{t}-\int\limits_1^{1+\sqrt 2}\mathrm dt\,\frac {\log 2t}t$$The second integral is trivial and numerically is equal as$$\mathfrak{I}_2=\int\limits_1^{1+\sqrt2}\mathrm dt\,\frac {\log 2t}t=\log 2\arcsinh1+\frac 12\arcsinh^21$$Now move on to the second integral. By integration by parts on taking $v=\log t$, the integral becomes$$\begin{align*}\mathfrak{I}_1 & =\log 2\arcsinh 1+\arcsinh^21+\frac 12\int\limits_1^{(1+\sqrt 2)^2}\mathrm dt\,\frac {\log t}{1-t}\\ & =\log 2\arcsinh 1+\arcsinh^21+\frac 12\sum\limits_{n\geq0}\int\limits_1^{(1+\sqrt 2)^2}\mathrm dt\, t^n\log t\end{align*}$$The last integral can be computed using a simple integration by parts. To save time, I used Wolfram Alpha, but you get the idea. Going off a detour$$\int\limits_1^{(1+\sqrt{2})^2}\mathrm dt\,t^n\log t=\frac 1{(n+1)^2}-\frac {(1+\sqrt2)^{2n+2}}{(n+1)^2}+\frac {2\arcsinh1(1+\sqrt2)^{2n+2}}{n+1}$$Summing and taking half gives the integral as$$\mathfrak{I}_1=\log 2\arcsinh1+\arcsinh^21+\frac {\pi^2}{12}-\frac 12\operatorname{Li}_2\left[(1+\sqrt2)^2\right]-\arcsinh1\log\left[1-(1+\sqrt2)^2\right]$$Putting everything together$$\int\limits_0^1\mathrm dx\,\frac {\log x}{\sqrt{1+x^2}}\color{blue}{=\frac 12\arcsinh^21+\frac {\pi^2}{12}-\frac 12\operatorname{Li}_2\left[(1+\sqrt2)^2\right]-\arcsinh1\log\left[1-(1+\sqrt2)^2\right]}$$Wolfram Alpha confirms numerically.

  • $\begingroup$ Thank you!,and what did you do a beautiful answer! $\endgroup$
    – FofX
    Sep 3, 2018 at 5:50
  • $\begingroup$ But I think,there is a typo?x=(t^2-1)/2t $\endgroup$
    – FofX
    Sep 3, 2018 at 6:03
  • $\begingroup$ @FofX You're right. It's fixed now. $\endgroup$
    – Frank W
    Sep 3, 2018 at 6:10

\begin{align} \int_0^1 x^\alpha\ dx &= \dfrac{x^{\alpha+1}}{\alpha+1}\\ \dfrac{d}{d\alpha}\int_0^1 x^\alpha\ dx &=\dfrac{d}{d\alpha}\dfrac{x^{\alpha+1}}{\alpha+1}\\ \int_0^1 x^\alpha\ln x\ dx &=\dfrac{x^{\alpha+1}((\alpha+1)\ln x-1)}{(\alpha+1)^2}\Big|_0^1=\dfrac{-1}{(\alpha+1)^2} \end{align} \begin{align} I &= \int_{0}^{1} \frac{\ln x}{\sqrt {1+x^2}}dx \\ &= \int_0^1\ln x\left(1-\dfrac12x^2+\dfrac38x^4+\cdots\right)\ dx\\ &= \int_{0}^{1} \frac{\ln x}{\sqrt {1+x^2}}dx \\ &= \int_0^1\left(\ln x-\dfrac12x^2\ln x+\dfrac38x^4\ln x+\cdots\right)\ dx \\ &= -1+\dfrac{1}{18}-\dfrac{3}{200}+\cdots \end{align}


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