hallo all,
I have the following problem:
Sorry, made an error and typed the wrong one.
$\displaystyle
\int
$ x^(-2 ). e^x dx is the one I mean.
Is it even possible to solve this?
hallo all,
I have the following problem:
Sorry, made an error and typed the wrong one.
$\displaystyle
\int
$ x^(-2 ). e^x dx is the one I mean.
Is it even possible to solve this?
Sure. But let's begin by converting that $\displaystyle \grave{o}$ symbol to the more familiar $\displaystyle \int$:
$\displaystyle \int x-2ex\;dx$
As written, the integral is simply performed:
$\displaystyle \int x-2ex\;dx=(1-2e)\int x\;dx=(\frac{1-2e}{2})x^2+C$
But I suspect that you actually mean this:
$\displaystyle \int x-2e^x\;dx$
For such an integral, we break it up into two portions:
$\displaystyle \int x\;dx-2\int e^x\;dx$
And hopefully you can take it from there.
That is correct, it has no nice indefinite form.
We can write it in terms of the Exponential Integral, though.
$\displaystyle \int\frac{e^{x}}{x^{2}}dx=\frac{-e^{x}}{x}-Ei(1,-x)$
Ei is the notation for the Exponential Integral.
$\displaystyle Ei(a,z)=\int_{x}^{\infty}\frac{e^{-t}}{t}dt$