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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.

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    $\begingroup$ Do you mean $e^{(e^{2015x})}$ or $(e^e)^{2015x}$? $\endgroup$ – Pim Feb 19 '15 at 6:19
  • $\begingroup$ To get an idea calculate the derivatives of $e^{e^x}$ and $e^{e^{e^x}}$. $\endgroup$ – Jyrki Lahtonen Feb 19 '15 at 6:23
  • $\begingroup$ Is this a question from an on-going contest? $\endgroup$ – Joel Reyes Noche Feb 19 '15 at 8:51
  • $\begingroup$ It's $2015$ alright... $\endgroup$ – AvZ Feb 19 '15 at 9:03
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Hint: Substitute $ u=e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x} $.

Then $$ du=2015e^{2015x+e^{e^{2015x}}+e^{2015x}}dx . $$

Therefore $$\int e^{e^{e^{e^{2015x}}}}e^{e^{e^{2015x}}}e^{e^{2015x}}e^{2015x}dx=\int \frac{ue^{u-e^{e^{2015x}}}}{2015}dx \dots $$

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    $\begingroup$ Are you sure that you differentiated $u$ correctly? It seems to me that the product/chain rules would give more terms. $\endgroup$ – Jyrki Lahtonen Feb 19 '15 at 6:26
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    $\begingroup$ Right idea, wrong substitution. $\endgroup$ – Mike Feb 19 '15 at 6:31
  • $\begingroup$ I edited my solution $\endgroup$ – ASB Feb 19 '15 at 6:53
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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

$$\frac1{2015}\int_{e^{e^{e^0}}}^{e^{e^{e^{2015(2015)}}}}e^{u_3}du_3=\frac1{2015}(e^{e^{e^{e^{2015^2}}}}-e^{e^e})$$

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

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