-1

Why is there no general form for the $$\int \frac{u}{v}$$

The idea why I thought about this is becausewe can differentiate a function of the form $u/v$ means its some other functions integral so there might be a remote probability that there is some way to get the integral of the form $u/v$.

Or might someone prove that there can’t be an integration done by some general form. Thanks. Guide me to other question if there exists such question.

Archis Welankar
  • 15,910
  • 1
    Was there a general form for $\int uv $? If there isn't, how can there be one for $\int \frac{u}{v}$? And if there is, isn't it easy to get $\int \frac{u}{v}$? – S.C.B. Mar 27 '16 at 05:36
  • u,v can be polynomials, trigonometric stuff, anything? – Nikunj Mar 27 '16 at 05:36
  • Wolframalpha says that there is no result found in terms of standard mathematical functions... – S.C.B. Mar 27 '16 at 05:40
  • 1
    @MXYMXY were you seriously expecting a general result? – Nikunj Mar 27 '16 at 05:42
  • @Nikunj Nope, I wasn't. I was trying to tell the OP there is no general known result. – S.C.B. Mar 27 '16 at 05:42
  • I will borrow a quote from one of our moderators, mixedmath "There is no generally easy way. For example, we know the antiderivative of both $\sin x$ and $x$, but there is no elementary antiderivative of their product $\frac{\sin x }{x}$" – S.C.B. Mar 27 '16 at 05:43
  • 1
    but $\int \frac{u}{v}$ can be worked out for most polynomials if you are willing to – Nikunj Mar 27 '16 at 05:45
  • Because $\frac d{dx}\colon \frac uv \mapsto \frac{vu' - uv'}{v^2}$, the quotient rule gives you a way to integrate stuff that looks like $\frac{vu' - uv'}{v^2}$ (which you've probably noticed, doesn't come up very often). There is no generally good way to write down the set of 'ingredients' $??$ for which $\frac d{dx}\colon ?? \mapsto \frac uv$. – pjs36 Mar 27 '16 at 05:46
  • No we have $\int u.v=u\int v+\int(u'\int v)$ – Archis Welankar Mar 27 '16 at 05:48
  • @ArchisWelankar So is it necessary that both the integrals work out? – Nikunj Mar 27 '16 at 05:48
  • No expecting there is so integration becomes easier – Archis Welankar Mar 27 '16 at 05:50
  • Do you mean "I am not expecting any general form. If there was, integration would become much easier."? – S.C.B. Mar 27 '16 at 05:52
  • Yes missed the t – Archis Welankar Mar 27 '16 at 06:03

1 Answers1

2

Your problem is similar to integrating $f(x)g(x)$. So here, I will leave you a link to this question and the accepted answer.

There is no generally easy way. For example, we know the antiderivative of both $\sin x$ and $\frac{1}{x}$, but there is no elementary antiderivative of their product $\frac{\sin x }{x}$.

Of course, as the second answer in the link points out, you could use integration by parts.

If $w=\frac{1}{v}$, then $$\int \frac{u}{v}=\int uw$$

From where you can proceed.

Community
  • 1
S.C.B.
  • 22,768