I've been having some trouble understanding a simplification involving logarithms that I came across in my notes. It is at the end of a partial fraction integration problem.
$$I = \int \frac{2}{x(2x-1)}dx $$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;I = \int \frac{2}{x(2x-1)} dx = \frac{A}{x} + \frac{B}{2x-1} $$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;I = \int \frac{2}{x(2x-1)} dx = \frac{A(2x-1)+Bx}{x(2x-1)} $$
$$ \;2 = A(2x-1) + Bx $$
$$ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x: 0 = 2A + B \;\;\;\;\;\;\;\;\;\;\;\;\;const: 2 = -A$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; A = -2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;B = 4 $$
$$\;\;\;\;\;\;\;\; I = \int \frac{-2}{x} + \frac{4}{2x-1} dx $$ $$ \;\;\;\;\;\;\;\;\;\;\;\;\;\;I = -2lnx + 2ln(2x-1) + c $$
Then this is where my lecturer wrote what something I don't understand. I realise he defined A earlier on in the problem but that makes the equation incorrect doesn't it?
$$ I = ln\left[A\left(\frac{2x-1}{x}\right)^2\right] $$
The next step is my simplification, which somebody said is incorrect as I used a log rule incorrectly. $$ I = lnx^{-2}+ln(2x-1)^2+c $$ $$ I = ln\left(\frac{(2x-1)^2}{(x)^2}\right) = ln\left(\frac{(2x-1)}{(x)}\right)^2+c $$
Is the following correct (this is the rule I used): $$-2ln(x) = ln(x^{-2}) = ln \left(\frac{1}{x^2}\right)$$
This is my first question here and it took me ages to write all this using mathjax. Is this the only way to write out equations here? Seems very laborious. Anyways, many thanks!
The rules that you use are correct.
But The final answer is true iff the problem was: $$\int\frac{2}{x(2x-1)}dx = \ln{\left[A\left(\frac{2x-1}{x}\right)^2\right]}$$
But if the problem be what you wrote, the answer is a quite different: $$\int\frac{2}{x(x-1)}dx = \ln{\left[A\left(\frac{x-1}{x}\right)^2\right]}$$
If it was a typo, you are right at all.
And about make constant $c$ to coefficient $A$. Since $\log$ is a onto function, there is $A$ that $c=\log A$.