Question Number 12703 by @ANTARES_VY last updated on 29/Apr/17
$$\boldsymbol{\mathrm{this}}\:\:\oint\left(\boldsymbol{\mathrm{x}}\right)=\frac{\boldsymbol{\mathrm{lnx}}^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\:\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{range}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{function}}. \\ $$
Answered by mrW1 last updated on 29/Apr/17
![t=ln x y=((2ln x)/(1+ln^2 x))=((2t)/(1+t^2 )) yt^2 −2t+y=0 D=(−2)^2 −4y^2 =4(1−y^2 )≥0 1−y^2 ≥0 y∈[−1,1]](Q12710.png)
$${t}=\mathrm{ln}\:{x} \\ $$$${y}=\frac{\mathrm{2ln}\:{x}}{\mathrm{1}+\mathrm{ln}^{\mathrm{2}} \:{x}}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${yt}^{\mathrm{2}} −\mathrm{2}{t}+{y}=\mathrm{0} \\ $$$${D}=\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}−{y}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\mathrm{1}−{y}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${y}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$