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So compute ##I_n:=\displaystyle{\int_\frac{1}{n}^n} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx##.archaic said:These are speculations, I can't really answer the question.
$$
S = \int_{\frac{1}{2}}^{3} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx \, + \, \int_{\frac{1}{3}}^{2} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx\\
= \int_{\frac{1}{3}}^{3} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx+\int_{\frac{1}{2}}^{2} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx
$$
This seemed to be equal to zero so I tried computing the area using some triangles and rectangles. I think that it's because of the fact that ##\log{\frac{1}{a}}+\log{a}=0## or something like this.