# integration of 1/x

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• Apr 19th 2014, 12:44 AM
delso
integration of 1/x
can i change the above expression to the below expression? why?

• Apr 19th 2014, 12:58 AM
delso
Re: integration of 1/x
here's my question actually https://www.flickr.com/photos/123101...3/13933620504/

here's the working, second photo is continuous from the first
https://www.flickr.com/photos/123101...3/13910042102/
https://www.flickr.com/photos/123101...3/13933180755/

if i cant do it in such way, can you please show me how do u get the ans please? thanks in advance!
• Apr 19th 2014, 07:50 AM
delso
Re: integration of 1/x
anybody can help plaese?
• Apr 19th 2014, 08:52 AM
romsek
Re: integration of 1/x
The first part is pretty straightforward

$y=Vx$

$\dfrac {dy}{dx}=V+x\dfrac {dV}{dx}$

just substitute those into your original equation and simplify.

Then the diff eq in V is separable so separate it and solve.
• Apr 19th 2014, 08:59 AM
JeffM
Re: integration of 1/x
Delso

For next time, it does not help to respond to your own initial post. That means your thread disappears from the list of unanswered posts and may greatly delay your getting an answer.
• Apr 19th 2014, 09:45 AM
Shakarri
Re: integration of 1/x
$\displaystyle \int 1+ \frac{6}{6-x} dx$

Well you are right that integration can be separated when two expressions are added or subtracted, integration has this additive property. The equation would become

$\displaystyle \int 1 dx + \int \frac{6}{6-x} dx$

But you made a little mistake with your fraction, the fraction can be changed to

$\displaystyle \int 1 dx + \int \frac{3}{\frac{1}{2}(6-x)} dx$

Changing the fraction doesn't really help with solving it though, try the change of variable y=6-x
• May 10th 2015, 03:16 PM
TheVirtualMathematician
Re: integration of 1/x