TITLE: How to evaluate $ \int_{0}^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx $?
QUESTION [6 upvotes]: I need to evaluate the following integral. 
$$\int_{0}^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx. $$
But i still  have no idea how to do it. Can anyone please give me a help? Thanks a lot.

REPLY [5 votes]: Repeatedly using the substitution $u=e^{2015x}$ gives
$$
\begin{align}
\int_0^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}\,\mathrm{d}x
&=\frac1{2015}\int_1^{e^{2015^2}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}\,\mathrm{d}x\\
&=\frac1{2015^2}\int_{e^{2015}}^{e^{2015e^{2015^2}}}e^{e^{2015x}}e^{2015x}\,\mathrm{d}x\\
&=\frac1{2015^3}\int_{e^{2015e^{2015}}}^{e^{2015e^{2015e^{2015^2}}}}e^{2015x}\,\mathrm{d}x\\
&=\frac1{2015^4}\int_{e^{2015e^{2015e^{2015}}}}^{e^{2015e^{2015e^{2015e^{2015^2}}}}}\,\mathrm{d}x\\
&=\frac1{2015^4}\left({e^{2015e^{2015e^{2015e^{2015^2}}}}}-e^{2015e^{2015e^{2015}}}\right)
\end{align}
$$