Reply 1
Have you thought about the +c?
Reply 2
Riiight so because it’s a constant it counts as + c, seems a little ambiguous though
Reply 3
Not sure what you mean, but there are a few ways you could think about it, so
•
Whats the derivative of ln(kx) - either chain rule or log property first.
•
The integration problem can be written as 1/k Int 1/x dx so ...
•
your way(?) was to write it as 1/k Int k/(kx) dx and if so when you write down the integral as ln(kx)/k + c, how can you rewrite the ln() part and what does that mean about the constant?
Reply 4
•
Whats the derivative of ln(kx) - either chain rule or log property first.
•
The integration problem can be written as 1/k Int 1/x dx so ...
•
your way(?) was to write it as 1/k Int k/(kx) dx and if so when you write down the integral as ln(kx)/k + c, how can you rewrite the ln() part and what does that mean about the constant?
You can write it as ln(x)/k + ln(k)/k + c right?
Reply 5
Yes which is
ln(x)/k + C
So its the "same". If you think about the first point above so the derivative of ln(kx) then
ln(kx) = ln(k) + ln(x)
so the derivatives of ln(x) and ln(kx) are the same as the functions have a constant difference ln(k).
Reply 6
•
Whats the derivative of ln(kx) - either chain rule or log property first.
•
The integration problem can be written as 1/k Int 1/x dx so ...
•
your way(?) was to write it as 1/k Int k/(kx) dx and if so when you write down the integral as ln(kx)/k + c, how can you rewrite the ln() part and what does that mean about the constant?
My way was to write it as (1/k)ln(kx)
Reply 7
I think that was the third point so note the integration problem is
1/k Int k/(kx) dx
where the integrand is f'(x)/f(x) which has an integral ln(f(x)) + c? Half guessing the result, its easier to rewrite the original integral problem as
1/k Int 1/x dx
Reply 8
ln(x)/k + C
So its the "same". If you think about the first point above so the derivative of ln(kx) then
ln(kx) = ln(k) + ln(x)
so the derivative of ln(x) and ln(kx) are the same as the functions have a constant difference ln(k). So the anti derivatives of 1/x and 1/(kx) differ by a constant which is absorbed by the +c part (which is important when you write the stuff in the OP)
Wait so I think I get it because if you the differentiate ln (x)/k + ln (k) + c you get 1/kx
Reply 9
Yes. Go through the 3 previous points and make sure you understand them. Is a small/subtle thing to be clear about. Either form (op) is correct, though ln(x)/k+c is the simplest so preferred.
Reply 10
Okay thank you very much
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