Integration of functions of form ##\dfrac{1}{ax+b}##

This is a bit confusing...conflicting report from attached wolfram and symbolab. Which approach is correct?

Symbolab seems correct to me and it is according to the standard definition of the function ##f(x)=ln(x)## which is defined only for real positive numbers ##x>0## and has derivative ##f'(x)=\frac{1}{x}##

Wolfram says in small letters "assuming a complex valued logarithm", which probably means it considers the analytic continuation of the function ##f(x)## so that it is defined also on negative real numbers. and that's why it doesn't put the (3x-1) into absolute value bars. So wolfram is probably correct too.

On why it has (3x-1) and not (6x-2) this isnt big problem, since it is ##\ln(6x-2)=\ln2+\ln(3x-1)##so it is a matter of constant, indefinite integrals can differ by a (additive) constant and be the same thing.

Delta2 said:

Symbolab seems correct to me and it is according to the standard definition of the function ##f(x)=ln(x)## which is defined only for real positive numbers ##x>0## and has derivative ##f'(x)=\frac{1}{x}##

Wolfram says in small letters "assuming a complex valued logarithm", which probably means it considers the analytic continuation of the function ##f(x)## so that it is defined also on negative real numbers. and that's why it doesn't put the (6x-2) into absolute value bars. So wolfram is probably correct too.

For wolfram the approach is as follows;

$$\int \dfrac{1}{6x-2} dx= \int \dfrac{1}{2(3x-1)} dx = \dfrac{1}{2}\int \dfrac{1}{(3x-1)} dx$$

Letting ##u =3x-1##

...

will lead to,

$$\dfrac{1}{2}\int \dfrac{1}{(3x-1)} dx = \dfrac{1}{6} \ln |3x-1|+c$$

Which seems to be correct.

Read again my update post #2 on why it doesn't matter the (3x-1) versus (6x-2) argument.

Delta2 said:

Read again my update post #2 on why it doesn't matter the (3x-1) versus (6x-2) argument.

Noted; so both responses are mathematically correct? Cheers man.

chwala said:

For wolfram the approach is as follows;

$$\int \dfrac{1}{6x-2} dx= \int \dfrac{1}{2(3x-1)} dx = \dfrac{1}{2}\int \dfrac{1}{(3x-1)} dx$$

Letting ##u =3x-1## will realize,

$$\dfrac{1}{2}\int \dfrac{1}{(3x-1)} dx = \dfrac{1}{6} \ln |3x-1|+c$$

Which seems to be correct.

That is correct, just note that wolfram uses the log function and doesn't put the argument into absolute bars and it seems to take the log function as the analytic continuation in negative numbers of the ln function (which is defined only in positive numbers).

chwala said:

Noted; so both responses are mathematically correct? Cheers man.

Both are correct but the wolfram answer is for calculus on $$\mathbb{C}$$ (the set of complex numbers) while Symbolab is for calculus on $$\mathbb{R}$$ (the set of real numbers).

The two answers are the same thing for ##6x-2>0\iff x>1/3## but for ##x<1/3## they are not the same thing.

chwala said:

Noted; so both responses are mathematically correct? Cheers man.

Yes, almost. ##\log |6x-2| +C = \log |3x-1| +\underbrace{\log 2 + C}_{=C'}.## The absolute value function should be included in both cases since I assume you were looking for the real logarithm and not for the complex one.

chwala said:

Homework Statement: How do we deal with the integration of $$\int \dfrac{1}{6x-2} dx?$$
Relevant Equations: Natural logs integration

This is a bit confusing...conflicting report from attached wolfram and symbolab. Which approach is correct?

For ##a\neq 0##,
[tex]
\int \frac{1}{ax+b}dx = \frac{1}{a}\int \frac{1}{ax+b}d(ax+b) = \frac{1}{a}\ln |ax+b| + C.
[/tex]

nuuskur said:

For ##a\neq 0##,
[tex]
\int \frac{1}{ax+b}dx = \frac{1}{a}\int \frac{1}{ax+b}d(ax+b) = \frac{1}{a}\ln |ax+b| + C.
[/tex]

As we speak of it. Here are tons of integrations:
https://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Integrale
It's not in English but the formulas don't care.

fresh_42 said:

It's not in English but the formulas don't care.

Yes the language of mathematics can be universal!