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Question Number 169825 by mathlove last updated on 10/May/22

Answered by Mathspace last updated on 10/May/22

$$I{=}_{x=\frac{1}{t}}-{\int }_0^{\mathrm{\infty }}\frac{ln\left(\frac{1+\frac{1}{t^{11}}}{1+\frac{1}{t^3}}\right)}{(1+\frac{1}{t^2})(-lnt)}(-\frac{dt}{t^2})$$$$=-{\int }_0^{\mathrm{\infty }}\frac{ln\left(t^{-8}\left(\frac{1+t^{11}}{1+t^3}\right)\right)}{(t^2+1)lnt}dt$$$$=-{\int }_0^{\mathrm{\infty }}\frac{ln\left(\frac{1+t^{11}}{1+t^3}\right)}{(t^2+1)lnt}dt+8{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+1}$$$$=-I+8.\frac{\pi }{2}=-I+4\pi \Rightarrow$$$$2I=4\pi \Rightarrow I=2\pi$$

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