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I've been trying for an embarrassingly long time to figure this one out. It looks like it should crack under integration by parts and integration by substitution, but I am having trouble with it. Any pointers?

$$\int {\exp(a \sqrt{x^2 + b} ) \over \sqrt{x^2+b} } dx$$

Ragib Zaman
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Mark Borgerding
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2 Answers2

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The special case $a=1,b=0$ simplifies to

$$\int \frac{e^x}{x} \, dx$$

Which is known as the Exponential Integral where no closed form is known. Therefore your integral has no general closed form either.

Note: By closed form I mean that it is not expressible in terms of elementary functions.

Listing
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  • ...I consider the exponential integral as a closed form expression in its own right. Better to say that the exponential integral is not expressible in terms of the elementary functions... – J. M. ain't a mathematician Nov 28 '11 at 13:05
  • I added a note to clarify what I mean, although if you consider $\text{Ei(x)}$ to be a closed form expression you could define every integral as a function and call it closed form :-) – Listing Nov 28 '11 at 13:09
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    Sure, but another part of my "closed-form" schtick is that there's a whole corpus of identities relating it to a lot of other functions. ;) You don't have that with, say, $\int u^u \mathrm du$. – J. M. ain't a mathematician Nov 28 '11 at 13:31
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The natural substitution is to let $u=a\sqrt{x^2+b}$ so that $du=\frac{ax}{\sqrt{x^2+b}}$ and $x^2=\frac{u^2}{a^2}-b$. Our integral then is

$$\int \frac{e^u}{\sqrt{u^2-ba^2}} du.$$ However, there is no way to deal with something of this form. Are you sure that you are not missing a multiple of $x$? In other words, I think the question should be to integrate $$\int \frac{x\exp(a\sqrt{x^2+b})}{\sqrt{x^2+b}}dx$$ because then it gives $$\frac{1}{a}e^{a\sqrt{x^2+b}}$$ as the anti derivative.

Eric Naslund
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