Question about logarithm in integral

Question

I haven't solved any problems with integrals for ages so I don't remember some basic things. Here's my problem (it's part of solving a differential equation):

$\int \frac{1}{y} \operatorname{dy} =\int \frac{(1-x)}{x}\operatorname{dx}$

My solution is this:

$\ln y=\int\frac{1}{x}-1 \operatorname{dx}$

$\ln y = \int\frac{1}{x}\operatorname{dx}-\int1 \operatorname{dx}$

$\ln y= \ln x-x+c$

But according to my book the solution is $\ln x-x+\ln c$

What I am missing?

Answer 1

set $$C'=\ln(c)$$ and that is your result

Answer 2

We have $$ (\ln x-x+c)'=\frac{1-x}x $$ and $$ (\ln x-x+\ln c)'=\frac{1-x}x $$ both antiderivatives are OK, just note that $c$ in the first identity is not the same constant as $c$ in the second one.

Answer 3

Given that $ln(y)= ln(x)- x+ c$, $y= e^{ln(x)- x+ c}= e^{ln(x)}(e^{-x})(e^c)= e^cxe^{-x}$. We could then decide to write $C= e^c$ to get $y= Cxe^{-x}$.

Or we could write $ln(y)= ln(x)- x+ ln(c)= e^{ln(x)-x+ ln(c)}= x(e^{-x})(e^{ln(c)})= cxe^{-x}$.

Either way, since "C" or "c" can be any number, we get the same answer.