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Question Number 12703 by @ANTARES_VY last updated on 29/Apr/17

$$\mathbf{this}\oint \left(\mathbf{x}\right)=\frac{{\mathbf{lnx}}^2}{1+{\mathbf{ln}}^2\mathbf{x}}\mathbf{find}\mathbf{the}$$$$\mathbf{range}\mathbf{of}\mathbf{the}\mathbf{function}.$$

Answered by mrW1 last updated on 29/Apr/17

$$t=\ln x$$$$y=\frac{2\ln x}{1+{\ln }^{2}x}=\frac{2t}{1+t^2}$$$${yt}^2-2t+y=0$$$$D={(-2)}^2-4y^2=4(1-y^2)⩾0$$$$1-y^2⩾0$$$$y\in [-1,1]$$

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