- #36

- 17,591

- 18,100

So compute ##I_n:=\displaystyle{\int_\frac{1}{n}^n} \dfrac{1}{\sqrt{x^2+1}}\,\dfrac{\log(x)}{\sqrt{x}}\,dx##.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.