Is it possible to find the integral of ##f(x)/x^2##?

[MevsEinstein]

TL;DR

What the title says

I am creating an integration technique and I have only one step left! I need to integrate $f(x)/x^2$ and then I'll be done. So I want to know if integrating this is possible.

Wolfram Alpha can't integrate it, but there are problems that it couldn't solve, so I'm not 100% sure that Wolfram Alpha is right.

 

[pbuk]

Oops

 

[DaveE]

You'll need to tell us about f(x) and the domain. Some functions just aren't integrable. For example, if $f(x)=x$, $\int_{-1}^1 \frac{1}{x} \, dx$ doesn't work. Wolfram Alpha told me so.

 

[mathman]

For arbitrary f(x), the answer is no.

 

[mfb]

You can't get an explicit solution without knowing what f(x) is. You can use integration by parts to express the integral in terms of integrals or derivatives of f(x). For some f(x) this can lead to simpler expressions, but that depends on f(x).

 

[jbriggs444]

So what if $f(x)$ has a domain that accepts all integers?

That does not narrow things down very much.

If you have a function that is integrable and you undefine it at all integers, it does not become unintegrable.

 

[pbuk]

So what if $f(x)$ has a domain that accepts all integers?

That doesn't help because:

You'll need to tell us about f(x)

For arbitrary f(x), the answer is no.

You can't get an explicit solution without knowing what f(x) is.

Not sure how many different ways we can say the same thing.

 

[TeethWhitener]

So what if $f(x)$ has a domain that accepts all integers?

What do you mean? Didn't @DaveE provide you with an example of an $f(x)$ that is defined for all $\mathbb{R}$?

 

[Office_Shredder]

If there was an easy way to integrate $f(x)/x^2$, then given $f$, you can just say $g(x)=x^2 f(x)$ and then compute the integral of $g(x)/x^2$. Then you have the integral of $f$.

So whatever other intermediate steps you were going to make, they wouldn't actually be needed.