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Question Number 167150 by mathlove last updated on 08/Mar/22

Answered by som(math1967) last updated on 08/Mar/22

$$let\left({arctanx}^{{arctanx}^{...}}\right)=z$$$${arctanx}^{{arctanx}^{...}}lnarctanx=lnz$$$$zlnarctanx=lnz$$$$\frac{dz}{dx}lnarctanx+\frac{z}{(1+x^2)arctanx}=\frac{1}{z}\frac{dz}{dx}$$$$\Rightarrow dz=\frac{z^2}{(1+x^2)arctanx}\times \frac{1}{(1-zlnarctanx)}dx$$$$=\frac{{\left({arctanx}^{{arctanx}^{...}}\right)}^{2}dx}{arctanx(1+x^2)(1-{arctanx}^{{arctanx}^{..}}lnarctanx)}$$$$\int dz=z+c$$$$={arctanx}^{{arctanx}^{...}}+c$$

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