How to evaluate $ \int_{0}^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx $? 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 some help? Thanks a lot.
 A: Hint: Substitute $ u=e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x} $.
Then $$ du=2015e^{2015x+e^{e^{2015x}}+e^{2015x}}dx . $$
Therefore $$\int e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx=\int \frac{ue^{u-e^{e^{2015x}}}}{2015}dx \dots $$
A: If you don't immediately see it, you can start with the substitution
$$u_1=e^{2015x},du_1=2015e^{2015x}$$
$$\int_0^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx=\frac1{2015}\int_1^{e^{2015^2}}e^{e^{e^{u_1}}}e^{e^{u_1}}e^{u_1}du_1$$
We went from a product of four things to a product of three. And we can keep going with $u_2=e^{u_1}$ and $u_3=e^{u_2}$ to bring it down to $\frac1{2015}\int e^{u_3}du_3$.  The hard part is the bounds.  If you want to work things back, though, you can see that $u_3=e^{e^{e^{2015x}}}$.  So we have
$$\frac1{2015}\int_{e^{e^{e^0}}}^{e^{e^{e^{2015(2015)}}}}e^{u_3}du_3=\frac1{2015}(e^{e^{e^{e^{2015^2}}}}-e^{e^e})$$
A: Using the substitution $u_1=e^{2015x}$, then repeatedly using $u_{k+1}=e^{u_k}$, gives
$$
\begin{align}
\int_0^{2015}e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}\,\mathrm{d}x
&=\frac1{2015}\int_1^{e^{2015^2}}e^{e^{e^{u_1}}}e^{e^{u_1}}e^{u_1}\,\mathrm{d}{u_1}\\
&=\frac1{2015}\int_e^{e^{e^{2015^2}}}e^{e^{u_2}}e^{u_2}\,\mathrm{d}{u_2}\\
&=\frac1{2015}\int_{e^e}^{e^{e^{e^{2015^2}}}}e^{u_3}\,\mathrm{d}{u_3}\\
&=\frac1{2015}\int_{e^{e^e}}^{e^{e^{e^{e^{2015^2}}}}}\,\mathrm{d}{u_4}\\
&=\frac1{2015}\left({e^{e^{e^{e^{2015^2}}}}}-e^{e^{e}}\right)
\end{align}
$$
