When is the constant C part of the function when integrating

Question

Lets take the following example:

$\int \frac{x-4}{(x-2)(x-3)}dx$

The result I get is $2\ln|x-2|-\ln|x-3|+C$

The result in my testbook is:

$\ln \frac{C(x-2)^2}{x-3}$

As $\ln \frac{C(x-2)^2}{x-3} = \ln(x-2)^2 - \ln|x-3| + \ln C$ and $\ln C$ can be every possible rational number, I assume that this is not wrong, but do I have to write it this way?

Answer 1

As you have indicated $$ \ln \frac{C(x-2)^2}{x-3} = \ln(x-2)^2 - \ln|x-3| + \ln C$$ Thus you do not have to change your answer to the compact form unless it is specifically required.

I would check the similar problem's answer in the back of the text book to see what format is used and follow the same format. Specially if it is a homework problem and is graded by a graduate assistant who accepts the solutions which are exactly like what is in the back of the book or in the solution manual.

Answer 2

$2\ln (x-2) -\ln(x-3)+\ln c=\ln (x-2)^2 -\ln(x-3)+\ln c=\ln\frac{c(x-2)^2}{x-3}$

so you are right ... and so is your textbook.... well done!

Answer 3

I think both answers they are wrong.

The right answer is $$\int\frac{x-4}{(x-2)(x-3)}dx=\ln(x-2)^2-\ln(3-x)+C_1$$ for $x<2$, $$\int\frac{x-4}{(x-2)(x-3)}dx=\ln(x-2)^2-\ln(3-x)+C_2$$ for $2<x<3$ and $$\int\frac{x-4}{(x-2)(x-3)}dx=\ln(x-2)^2-\ln(x-3)+C_3$$ for $x>3$.