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Below is the integral I'd like to solve $ \int e^{e^{2016x}+6048x} \,dx $ or more general

$$ \int e^{e^{kx}+3kx} \,dx $$

I tried substituting $t = e^{kx}+3kx$, but can't solve $\int \frac{e^t}{(e^{kx} + 3)}\frac{dt}{k}$. I can't seem to get rid of the $x$ with a substitution.

Note: This question was posted earlier (minus the generalization to $k$) and while I was working on my solution it was deleted (only a few minutes after its posting). I don't remember the original author to give credit to. Also I couldn't find any duplicates and I think it's an interesting integral so I thought I'd post it with my solution. However, I welcome any other solutions, especially those more elegant, shorter, interesting, and/or clever than mine.

Dando18
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1 Answers1

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For the integral,

$$ J = \int e^{e^{kx}+3kx} \,dx $$

make the substitution $t \mapsto e^{kx}$

$$ J = \frac{1}{k} \int e^t t^2 \,dt $$

Now IBP where $u = t^2$ and $dv = e^t \,dt$

$$ J = \frac{1}{k} e^t t^2 - \frac{2}{k} \int e^t t \,dt $$

again IBP with $f = t$ and $dg = e^t \, du$

$$ J = -\frac{2}{k} e^t t + \frac{1}{k} e^t t^2 + \frac{2}{k} e^t + c$$

factor out $\frac{e^t}{k}$ and substitute back $t \mapsto e^{kx}$

$$ J = \frac{1}{k}e^{e^{kx}} \left(-2e^{kx} + e^{2kx} + 2 \right) + c $$

Dando18
  • 5,472
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    Interesting that you need special functions if $3$ is replaced with any real number besides a positive integer. – user49640 Jun 15 '17 at 02:54
  • interesting indeed. I guess the there's no way to calculate $\int e^u u^{n-1} , du$ if $n<1$ or $n \notin \mathbb{N}$ without special functions. I think this comes from the incomplete gamma representation shown here. – Dando18 Jun 15 '17 at 03:14