Integral exercise asked by Kunihiko Chikaya

Exercise:

Compute, $\large{\int_{{{e}^{{{e}^{e}}}}}^{{{e}^{{{e}^{{{e}^{e}}}}}}}{\frac{dx}{x\ln x.\ln \left( \ln x \right).\ln \left( \ln \left( \ln x \right) \right)}}}$

Solution: Let $w=\ln \left( \ln \left( \ln \left( \ln x \right) \right) \right)$ and take $u\left( x \right)=\ln \left( \ln \left( \ln x \right) \right)$

$v\left( x \right)=\ln \left( \ln x \right)\,\,\,\And \,\,\,t\left( x \right)=\ln x$

Thus $dw=\frac{u'\left( x \right)}{u\left( x \right)}dx$ , $u'\left( x \right)=\frac{v'\left( x \right)}{v\left( x \right)}\,\,,\,\,v'\left( x \right)=\frac{t'\left( x \right)}{t\left( x \right)}\,\,\,\,\And \,\,\,t'\left( x \right)=\frac{1}{x}$

$\Rightarrow dw=\frac{v'\left( x \right)}{v\left( x \right)u\left( x \right)}dx=\frac{t'\left( x \right)}{t\left( x \right)v\left( x \right)u\left( x \right)}dx=\frac{dx}{xt\left( x \right)v\left( x \right)u\left( x \right)}=\frac{dx}{x\ln x.\ln \left( \ln x \right).\ln \left( \ln \left( \ln x \right) \right)}$

But $w\left( {{e}^{{{e}^{{{e}^{e}}}}}} \right)=\ln \left( \ln \left( \ln \left( \ln {{e}^{{{e}^{{{e}^{e}}}}}} \right) \right) \right)=\ln \left( \ln \left( \ln {{e}^{{{e}^{e}}}} \right) \right)=\ln \left( \ln {{e}^{e}} \right)=\ln \left( e \right)=1$

And $w\left( {{e}^{{{e}^{e}}}} \right)=\ln \left( \ln \left( \ln \left( \ln {{e}^{{{e}^{e}}}} \right) \right) \right)=\ln \left( \ln \left( \ln {{e}^{e}} \right) \right)=\ln \left( \ln e \right)=\ln \left( 1 \right)=0$

So $\large{\int_{{{e}^{{{e}^{e}}}}}^{{{e}^{{{e}^{{{e}^{e}}}}}}}{\frac{dx}{x\ln x.\ln \left( \ln x \right).\ln \left( \ln \left( \ln x \right) \right)}=}\int_{0}^{1}{\frac{x\ln x.\ln \left( \ln x \right){{e}^{w}}dw}{x\ln x.\ln \left( \ln x \right).{{e}^{w}}}}=\int_{0}^{1}{dw}=1}$