# Thread: integration problem

1. ## integration problem

hallo all,

I have the following problem:

Sorry, made an error and typed the wrong one.
$
\int
$
x^(-2 ). e^x dx is the one I mean.

Is it even possible to solve this?

2. Originally Posted by Coele
hallo all,

I have the following problem:

òx-2·ex dx = ?

Is it even possible to solve this?
Sure. But let's begin by converting that $\grave{o}$ symbol to the more familiar $\int$:

$\int x-2ex\;dx$

As written, the integral is simply performed:

$\int x-2ex\;dx=(1-2e)\int x\;dx=(\frac{1-2e}{2})x^2+C$

But I suspect that you actually mean this:

$\int x-2e^x\;dx$

For such an integral, we break it up into two portions:

$\int x\;dx-2\int e^x\;dx$

And hopefully you can take it from there.

3. sorry,

my bad,

I made an error:

I ment this one:

e^x / x^2 dx (or as I wrote it before: x^-2 . e^x dx

4. Originally Posted by Coele
sorry,

my bad,

I made an error:

I ment this one:

e^x / x^2 dx (or as I wrote it before: x^-2 . e^x dx
I don't think this has an elementary, closed form antiderivative.

you could use the infinite series for $y = e^x$, divide by $x^2$, and integrate term for term.

5. That is correct, it has no nice indefinite form.

We can write it in terms of the Exponential Integral, though.

$\int\frac{e^{x}}{x^{2}}dx=\frac{-e^{x}}{x}-Ei(1,-x)$

Ei is the notation for the Exponential Integral.

$Ei(a,z)=\int_{x}^{\infty}\frac{e^{-t}}{t}dt$

6. I see

thanks a lot guys.