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How to integrate 1 over 5x?

How do you calculate:

∫15x dx \int \frac{1}{5x}\ dx

I thought that it would be 15ln(5x) \frac{1}{5} ln(5x) but Wolfram Alpha says the answer is 15ln(x) \frac{1}{5} ln(x) .

I understand that ln(5x) ln (5x) integrates to 1x \frac{1}{x} , as does ln(x) ln(x) . So how do I know which 1x \frac{1}{x} integrates to?
You take out the (1/5) to the front of the integral as it is a constant. Now you are simply integrating 1/x which is Lnx, now remember the constant so it is (1/5)Lnx

The integral of Ln5x is actually 5/(5x) which cancels down to 1/x
15ln∣x∣ \frac {1}{5} ln|x|
I don't really understand what you are trying to say but in that integral the 1/5 multiplying is not affected by the integral. So then you have 1/5*int(1/x). the integral of 1/x is ln(X) leaving you with 1/5In(x)+c. You can make sure by deriving.
Original post by maxcrichton-s
I don't really understand what you are trying to say but in that integral the 1/5 multiplying is not affected by the integral. So then you have 1/5*int(1/x). the integral of 1/x is ln(X) leaving you with 1/5In(x)+c. You can make sure by deriving.


I'm confused that both 15ln(5x)\frac{1}{5} ln(5x) and 15ln(x)\frac{1}{5} ln(x) differentiate to the same thing. Therefore why is integrating 1x \frac{1}{x} to 15ln(5x)\frac{1}{5} ln(5x) wrong?
Original post by xy_
I'm confused that both 15ln(5x)\frac{1}{5} ln(5x) and 15ln(x)\frac{1}{5} ln(x) differentiate to the same thing. Therefore why is integrating 1x \frac{1}{x} to 15ln(5x)\frac{1}{5} ln(5x) wrong?

They don't differentiate to the same thing. 15ln(5x)\frac{1}{5} ln(5x) differentiates to 1/x and 15ln(x)\frac{1}{5} ln(x) differentiates to 1/(5x)
Original post by xy_
I'm confused that both 15ln(5x)\frac{1}{5} ln(5x) and 15ln(x)\frac{1}{5} ln(x) differentiate to the same thing. Therefore why is integrating 1x \frac{1}{x} to 15ln(5x)\frac{1}{5} ln(5x) wrong?


Your answer is correct but it is missing +c. This is important because ∫15x.dx=15ln⁡∣5x∣+c\displaystyle \int \frac{1}{5x} .dx = \frac{1}{5}\ln |5x| +c

But note that ln⁡∣5x∣=ln⁡5+ln⁡∣x∣\ln |5x| = \ln5 + \ln |x| so what you ACTUALLY have is ∫15x.dx=15ln⁡∣x∣+15ln⁡5+c\displaystyle \int \frac{1}{5x} .dx = \frac{1}{5}\ln |x| + \frac{1}{5}\ln 5 +c

Now since 15ln⁡5\frac{1}{5}\ln 5 is a constant, we have 15ln⁡5+c=C=const. \frac{1}{5}\ln 5 +c = C = \text{const.} and we end up with the answer of ∫15x.dx=15ln⁡∣x∣+C\displaystyle \int \frac{1}{5x} .dx = \frac{1}{5}\ln |x| + C
Reply 7
Avatar for trythis
trythis
11
∫15x dx=15∫1x dx=15ln⁡(x)+c \int \frac{1}{5x}\ dx = \frac{1}{5} \int \frac{1}{x}\ dx = \frac{1}{5} \ln (x) + c

Note:
dydx (ln⁡x)=1x \frac{dy}{dx}\ (\ln x)= \frac{1}{x}
and:
dydx (ln⁡(5x))=dydx (ln⁡(5)+ln⁡(x))=1x \frac{dy}{dx}\ (\ln (5x)) = \frac{dy}{dx}\ (\ln (5) + \ln (x)) = \frac{1}{x}
(edited 6 years ago)
Reply 8
Avatar for Zaydy3
Zaydy3
11
its clearly 1/5 ln (5x) how are people getting 1/5 ln(x) xD
Original post by Zaydy3
its clearly 1/5 ln (5x) how are people getting 1/5 ln(x) xD


The answers are same, it's just the constants that differ, and they are easily dealt with. Look at my post above for clarification.
well it can be 1/5 ln |5x| + c
but then again, ln |5x| = ln 5 + ln |x|
so it's 1/5 ln |x| + 1/5 * ln 5 + c where you can say 1/5 * ln 5 + c is a constant so you get + c


=======

Too late :c RDK games already answered the question
(edited 6 years ago)
Reply 11
Avatar for Zaydy3
Zaydy3
11
Original post by RDKGames
The answers are same, it's just the constants that differ, and they are easily dealt with. Look at my post above for clarification.



oh thanks :smile:
Reply 12
Avatar for eemi
eemi
1
ln(5x) is 1/x but you forgot the 1/5 from the front
Original post by eemi
ln(5x) is 1/x but you forgot the 1/5 from the front

Please don't reply to posts from over a year ago.
Reply 14
Avatar for eemi
eemi
1
lul
Original post by DFranklin
Please don't reply to posts from over a year ago.

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