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Question Number 143046 by 0731619 last updated on 09/Jun/21

Answered by mindispower last updated on 09/Jun/21

$$x^{x^x}=f\left(x\right)\Rightarrow ln\left(fx\right)=f\left(x\right)ln\left(x\right)$$$$ln\left(f\left(x\right)\right)e^{-ln\left(f\left(x\right)\right)}=ln\left(x\right)$$$$\Rightarrow -ln\left(f\left(x\right)\right)=W(-ln\left(x\right))$$$$f\left(x\right)=e^{-W(-ln\left(x\right))}=\frac{W(-\mathit{l}\mathit{n}\left(\mathit{x}\right))}{-\mathit{l}\mathit{n}\left(\mathit{x}\right)}$$$$\int \frac{e^{-2\mathit{W}(-ln\left(x\right))}}{x(1-ln\left(x\right)e^{-W(-ln\left(x\right))})}$$$$ln\left(x\right)=y$$$$=\int \frac{e^{-2W(-y)}}{(1-{ye}^{-W(-y)})}dy$$$$\int \frac{e^{-W(-y)}}{e^{W(-y)}-y}dy$$$$W\left(y\right)e^{W\left(y\right)}=y$$$$\Rightarrow W^{\prime }\left(y\right)(1+W\left(y\right))e^{W\left(y\right)}=1$$$$W^{\prime }\left(y\right)=\frac{e^{-W\left(y\right)}}{1+W\left(y\right)}$$$$e^{W(-y)}-y=f\left(y\right),f^{\prime }\left(y\right)$$$$letg\left(y\right)=e^{W(-y)}$$$$g^{\prime }\left(y\right)=-W^{\prime }(-y)e^{W(-y)}=\frac{-1}{1+W(-y)}$$$$=\frac{-e^{W(-y)}}{e^{W(-y)}-y}\Rightarrow dg\left(y\right)e^{W(-y)}=\frac{dy}{e^{W(-y)}-y}$$$$\int \frac{1}{g^2}dg=-\frac{1}{g}+c$$$$=-e^{-W(-y)}+c$$$$=-e^{-W(-ln\left(x\right))}+c=\int \frac{x^{x^{..{.}^2}}}{x(1-x^{x^{x^{.}}}ln\left(x\right)}dx$$$$$$$$$$$$$$

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