How to evaluate the integral $\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx$ $$I=\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx$$
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.
 A: $\newcommand{\arcsinh}{\operatorname{arcsinh}}$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.
A: \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}
