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Question-167150




Question Number 167150 by mathlove last updated on 08/Mar/22
Answered by som(math1967) last updated on 08/Mar/22
 let (arctanx^(arctanx^(...) ) )=z   arctanx^(arctanx^(...) ) lnarctanx=lnz   zlnarctanx=lnz   (dz/dx)lnarctanx+(z/((1+x^2 )arctanx))=(1/z)(dz/dx)   ⇒ dz =(z^2 /((1+x^2 )arctanx)) ×(1/((1−zlnarctanx)))dx  =(((arctanx^(arctanx^(...) ) )^2 dx)/(arctanx(1+x^2 )(1−arctanx^(arctanx^(..) ) lnarctanx)))  ∫dz=z+c  =arctanx^(arctanx^(...) ) +c
$$\:{let}\:\left({arctanx}^{{arctanx}^{…} } \right)={z} \\ $$$$\:{arctanx}^{{arctanx}^{…} } {lnarctanx}={lnz} \\ $$$$\:{zlnarctanx}={lnz} \\ $$$$\:\frac{{dz}}{{dx}}{lnarctanx}+\frac{{z}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){arctanx}}=\frac{\mathrm{1}}{{z}}\frac{{dz}}{{dx}} \\ $$$$\:\Rightarrow\:{dz}\:=\frac{{z}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){arctanx}}\:×\frac{\mathrm{1}}{\left(\mathrm{1}−{zlnarctanx}\right)}{dx} \\ $$$$=\frac{\left({arctanx}^{{arctanx}^{…} } \right)^{\mathrm{2}} {dx}}{{arctanx}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{arctanx}^{{arctanx}^{..} } {lnarctanx}\right)} \\ $$$$\int{dz}={z}+{c} \\ $$$$={arctanx}^{{arctanx}^{…} } +{c} \\ $$

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