Hi all

Are there common ways to manipulate the integral

$\displaystyle \int\frac{fg'}{g}dx$

so that it can be evaluated? would one need to know the exact functions in question or can this be answered generally?

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- Feb 7th 2010, 04:24 AMfactfinderintegrate (f*g')/g
Hi all

Are there common ways to manipulate the integral

$\displaystyle \int\frac{fg'}{g}dx$

so that it can be evaluated? would one need to know the exact functions in question or can this be answered generally? - Feb 7th 2010, 04:44 AMMoo
Hello,

Nope it can't be solved without further information. It's not a known form :) - Feb 7th 2010, 08:15 AMchisigma
Calling $\displaystyle x$ the independent variable is...

$\displaystyle \frac{g^{'}(x)}{g(x)}= \frac{d}{dx} \ln g(x)$ (1)

... so that integrating by parts we have the identity...

$\displaystyle \int f(x)\cdot \frac{g^{'}(x)}{g(x)}\cdot dx = f(x)\cdot \ln g(x) - \int f^{'}(x)\cdot \ln g(x)\cdot dx$ (2)

Honestly I don't know if (2) can be of some utility for You (Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Feb 7th 2010, 11:05 PMchisigma
An interesting possibility is when is $\displaystyle f(x)= g(x) + \chi$ with $\displaystyle \chi$ an arbitrary constant, so that is $\displaystyle f^{'} (x) = g^{'} (x)$. In such a case is...

$\displaystyle \int f(x)\cdot \frac {g^{'} (x)}{g(x)}\cdot dx= 2\cdot g(x)\cdot \ln g(x) + (\chi -1)\cdot g(x) + c$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$