Integral for nested powers

Exercise:

Integrate, $\int{{{3}^{x}}{{3}^{{{3}^{x}}}}{{3}^{{{3}^{{{3}^{x}}}}}}{{3}^{{{3}^{{{3}^{{{3}^{x}}}}}}}}}dx$

Solution: Let $u={{3}^{x}}\Leftrightarrow du={{3}^{x}}\ln 3\,dx=u\ln 3\,dx\Leftrightarrow dx=\frac{du}{u\ln 3}$

So $\int{{{3}^{x}}{{3}^{{{3}^{x}}}}{{3}^{{{3}^{{{3}^{x}}}}}}{{3}^{{{3}^{{{3}^{{{3}^{x}}}}}}}}dx}=\int{u{{3}^{u}}{{3}^{{{3}^{u}}}}{{3}^{{{3}^{{{3}^{u}}}}}}\frac{du}{u\ln 3}}=\frac{1}{\ln 3}\int{{{3}^{u}}{{3}^{{{3}^{u}}}}{{3}^{{{3}^{{{3}^{u}}}}}}du}$

Let $w={{3}^{u}}\Leftrightarrow dw={{3}^{u}}\ln 3\,du\Leftrightarrow dw=w\ln 3\,du\Leftrightarrow du=\frac{dw}{w\ln 3}$

So $\frac{1}{\ln 3}\int{{{3}^{u}}{{3}^{{{3}^{u}}}}{{3}^{{{3}^{{{3}^{u}}}}}}}du=\frac{1}{{{\ln }^{2}}3}\int{{{3}^{w}}{{3}^{{{3}^{w}}}}dw}$

Let $v={{3}^{w}}\Leftrightarrow dv={{3}^{w}}\ln 3\,dw\Leftrightarrow dv=v\ln 3\,dw\Leftrightarrow dw=\frac{dv}{v\ln 3}$

So $\frac{1}{{{\ln }^{2}}3}\int{{{3}^{w}}{{3}^{{{3}^{w}}}}dw}=\frac{1}{{{\ln }^{3}}3}\int{{{3}^{v}}dv}$

In a similar way put $t={{3}^{v}}\Leftrightarrow dt={{3}^{v}}\ln 3dv\Leftrightarrow dv=\frac{dt}{t\ln 3}$

So $\frac{1}{{{\ln }^{3}}3}\int{{{3}^{v}}dv}=\frac{1}{{{\ln }^{4}}3}\int{dt=\frac{t}{{{\ln }^{4}}3}}+c$

Thus $\int{{{3}^{x}}{{3}^{{{3}^{x}}}}{{3}^{{{3}^{{{3}^{x}}}}}}{{3}^{{{3}^{{{3}^{{{3}^{x}}}}}}}}}dx=\frac{1}{{{\ln }^{4}}3}{{3}^{{{3}^{{{3}^{x}}}}}}+c$