How to evaluate :$$I=\int^1_0 \ln\left({x+\sqrt{1+x^2}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{x}}\right)dx$$
My attempt: \begin{align*} I & =\int^1_0 \ln\left({x+\sqrt{1+x^2}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{x}}\right)dx \\[2mm] & = \int^1_0 \ln\left({x+\sqrt{1+x^2}}\right)\ln\left({\frac{x}{1-\sqrt{1-x^2}}}\right)dx \end{align*} Hence: $$2I=\int^1_0 \ln\left({x+\sqrt{1+x^2}}\right)\ln\left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx$$
How do I continue to answer?
Too Long to be a comment: $$ \begin{aligned} \int_{0}^{1} \frac{\sqrt{1+x^2}-1}{x\sqrt{1-x^2}} \mathrm{d}x & = \int_{0}^{1} \frac{x}{(\sqrt{1+x^2}+1)\sqrt{1-x^2}} \mathrm{d}x \\ & = \frac1{2} \int_{0}^{1} \frac{\mathrm{d}x}{(\sqrt{1+x}+1)\sqrt{1-x}} \\ & = \frac1{2} \int_{0}^{1} \frac{\cos x}{\cos x+\sqrt{1-\sin x}} \mathrm{d}x\\ & = \frac1{2} \int_{0}^{\pi/2} \frac{\cos^2\!\frac{x}{2}-\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}-\sin^2\!\frac{x}{2}+\cos\frac{x}{2}-\sin\frac{x}{2}} \mathrm{d}x \\ & = \frac1{2} \int_{0}^{\pi/2} \mathrm{d}x - \frac1{2} \int_{0}^{\pi/2} \frac{\mathrm{d}x}{\sin\frac{x}{2}+\cos\frac{x}{2}+1} \\ & = \frac{\pi}{4} - \frac1{2} \int_{0}^{\pi/2} \frac{\mathrm{d}x}{2\sin\frac{x}{4}\cos\frac{x}{4}+2\cos^2\!\frac{x}{4}} \\ & = \frac{\pi}{4} - \int_{0}^{\pi/2} \frac{\mathrm{d}\tan\frac{x}{4}}{\tan\frac{x}{4}+1} \\ & = \frac{\pi}{4}-\ln\left(1+\tan\frac{\pi}{8}\right) = \frac{\pi}{4}-\frac{\ln2}{2} \end{aligned} $$
\begin{align} &\int^1_0 \ln\left({x+\sqrt{1+x^2}}\right)\ln{\frac{1+\sqrt{1-x^2}}{x}}\ dx\\ =& \int^1_0 \sinh^{-1}x\ \cosh^{-1}\frac1x\ \overset{ibp} {dx}\\ =&\int_0^1 \frac{\sinh^{-1}x}{\sqrt{1-x^2}}dx -\int_0^1 \frac{x\cosh^{-1}\frac1x}{\sqrt{1+x^2}}dx\\ =& \int_0^1\int_0^1 \frac x{\sqrt{(1-x^2)(1+x^2y^2)}}dy\ dx\\ &\>\>\>\>\>\>\>\>-\int_0^1 {\cosh^{-1}\frac1x}\ \overset{ibp} d\left(\sqrt{1+x^2}-1\right)\\ =&\int_0^1 \frac{\tan^{-1}y}{y}dy -\int_0^1 \frac{\sqrt{1+x^2}-1}{x\sqrt{1-x^2}}dx\\ =& \ G -\left(\frac\pi4 -\frac12\ln2\right) \end{align}
Integration by parts yields $$ \int \ln \left(x+\sqrt{1+x^2}\right) d x=x \ln \left(x+\sqrt{1+x^2}\right)-\sqrt{1+x^2}+C \\ \textrm{ and }\\ \frac{d}{d x} \ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)=-\frac{1}{x \sqrt{1-x^2}} \tag*{} $$ Using integration by parts again gives $$ \begin{aligned} & \int_0^1 \ln \left(x+\sqrt{1+x^2}\right) \ln \left(\frac{1+\sqrt{1-x^2}}{x}\right) d x \\ = & \int_0^1 \ln \left(\frac{1+\sqrt{1-x^2}}{x}\right) d\left[x \ln \left(x+\sqrt{1+x^2}\right)-\left(\sqrt{1+x^2}-1\right)\right]\\=&\int_0^1 \frac{\ln \left(x+\sqrt{1+x^2}\right)}{\sqrt{1-x^2}} d x-\int_0^1 \frac{\sqrt{1+x^2}-1}{x \sqrt{1-x^2}} d x\\=& \int_0^1 \frac{\sinh ^{-1} x}{\sqrt{1-x^2}}-\left(\frac{\pi}{4}-\frac{\ln 2}{2}\right) (\textrm{ Via } x=\sin \theta )\\=&G-\frac{\pi}{4}+\frac{\ln 2}{2} \end{aligned} $$ where $G=\int_0^1 \frac{\sinh ^{-1} x}{\sqrt{1-x^2}}dx$ is the Catalan’s constant.
Wish it helps.
Rewrite the logarithms as hyperbolic functions, then exploit Fubini's theorem:
$$\newcommand{arsinh}{\operatorname{arsinh}} \begin{align*} I &= \int_0^1 \arsinh x \operatorname{arsech}x \, dx \\ &= \int_0^1 \int_x^1 \frac{\arsinh x}{y\sqrt{1-y^2}} \, dy \, dx = \int_0^1 \int_0^y \cdots dx\,dy \\ &= \int_0^1 \frac{1 - \sqrt{1+y^2} + y \arsinh y}{y\sqrt{1-y^2}} \, dy & \text{by parts} \\ &= \int_0^1 \frac{1-\sqrt{1+y^2}}{y\sqrt{1-y^2}} \, dy + \int_0^1 \frac{\arsinh y}{\sqrt{1-y^2}} \, dy \end{align*}$$
A variety of trig/hyperbolic substitutions can be applied to the first integral to show
$$\begin{align*} \int \frac{1-\sqrt{1+y^2}}{y\sqrt{1-y^2}}\,dy &= \frac12 \operatorname{arsech}y^2 - \frac12 \arcsin y^2 - \operatorname{arsech} y + C \\ \implies \int_0^1 \cdots dy &= \frac12\log2 - \frac\pi4 \end{align*}$$
The latter integral is a known form of Catalan's constant.
