# Math Help - Different forms of same integral?

1. ## Different forms of same integral?

Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral

$\int \frac{1}{3x}dx$

Normally I would do this as following

$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$

However, is it not true that I can also make this of the form

$\int \frac{f'(x)}{f(x)}dx$

via

$\frac{1}{3}\int \frac{3}{3x}dx$

which via a simple u sub yields

$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$

Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?

$y=\frac{1}{3}\ln{x}+k$ and $\frac{1}{3}\ln{3x}+j$

as explicit functions of x?

Ta muchly.

2. Originally Posted by Xei
Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral

$\int \frac{1}{3x}dx$

Normally I would do this as following

$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$

However, is it not true that I can also make this of the form

$\int \frac{f'(x)}{f(x)}dx$

via

$\frac{1}{3}\int \frac{3}{3x}dx$

which via a simple u sub yields

$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$

Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?

Mr F says: They are equivalent. j + ln 3 is just as arbitrary as k .....

$y=\frac{1}{3}\ln{x}+k$ and $\frac{1}{3}\ln{3x}+j$

as explicit functions of x?

Mr F says: They already are explicit functions of x. Do you mean how to express them as explicit functions of y ....?

Ta muchly.
..

3. Originally Posted by Xei
Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral

$\int \frac{1}{3x}dx$

Normally I would do this as following

$\frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$

However, is it not true that I can also make this of the form

$\int \frac{f'(x)}{f(x)}dx$

via

$\frac{1}{3}\int \frac{3}{3x}dx$

which via a simple u sub yields

$\frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$

Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?
Since j and k are just arbitrary constants one can be set equal to the other.

By the way, just so that you are aware...
$\int \frac{1}{x}~dx = ln|x| + C$

-Dan

4. Yeah it's all right, I just wasn't bothered with the mod bit.

I was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.

Is it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?

Mr F: Yeah, y please. As simple as it'll go.

5. Hello, Xei!

You are basically correct . . .

$\int \frac{1}{3x}\,dx \;=\;\frac{1}{3}\int \frac{1}{x}\,dx\;=\;\frac{1}{3}\ln{x}+K$ .[1]

$\frac{1}{3}\int\frac{3}{3x}\,dx \;=\;\frac{1}{3}\ln{3x}+J$ .[2]

Is it the case that both are valid, but the constants $K$ and $J$ are different? .
. . . yes

Is $K\:=\:J+\ln3$? .
. . . um, not quite

From [2], we have: . $\frac{1}{3}\ln(3x) + J \;=\;\frac{1}{3}\left[\ln(3) + \ln(x)\right] + J \;=\;\frac{1}{3}\ln(3) + \frac{1}{3}\ln(x) + J$

. . $= \;\frac{1}{3}\ln(x) + \underbrace{\frac{1}{3}\ln(3) + J}_{\text{a constant}} \;=\;\frac{1}{3}\ln(x) + K$

6. Originally Posted by Xei
Yeah it's all right, I just wasn't bothered with the mod bit.

I was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.

Is it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?

Mr F: Yeah, y please. As simple as it'll go.
$y=\frac{1}{3}\ln{|x|}+k \Rightarrow 3y - 3k = \ln |x| \Rightarrow e^{3y - 3k} = x$

$\Rightarrow x = e^{3y}\, e{-3k} = A e^{3y}$.