Integrating Ln x: Does the Answer Always Equal 1/x?

[Steel_City82]

When you intergrate a Ln ax problem does is the answer always 1/x

for example integrate ln 2x
does it equal 1/x or ln 3x and so on

For some reason when i work it out on paper or my calculator it comes out to 1/x and I just don't think its right. I think I am just being catious since the problems I am doing will determine my grade for this semester.

 

[hypermorphism]

Note that by definition,
\ln x = \int_1^x \frac{1}{t} dt
While the derivative of ln(x) is 1/x, its integral certainly is not. you can find its integral by parts (product rule). If you want the derivative of ln(f(x)) for some function f, use the chain rule.
This will tell you that the derivative of ln(ax) for any non-zero constant a is indeed 1/x. What are your reasons for thinking this derivative to be incorrrect ?
This should just remind you that ln(ax) = ln(a) + ln(x), a simple algebraic rule of all logarithms.

 

[Steel_City82]

Well when you differentiate ln x will it always be 1/x even if its ln 2x or ln 3x.

the problem is an integration by parts and i set my u as ln 3x and my dv as 2x^4

when i try to find the answer for the du i get 1/x with my calc and also when i do it on paper. I just want to make sure that it correct.

 

[hypermorphism]

Well when you differentiate ln x will it always be 1/x even if its ln 2x or ln 3x.
the problem is an integration by parts and i set my u as ln 3x and my dv as 2x^4
when i try to find the answer for the du i get 1/x...

There's no problem there. Just keep going by parts until you get something you can work with.

 

[Steel_City82]

thanks for your help

 

[Jameson]

If you want to integrate ln(x), make u=ln(x) and dv=1. The answer is quite simple from there.

For any function in the form of aln(ax), the derivative will always be 1/x.