Evaluate $-\int \frac{dy}{y}$ $$-\int \frac{dy}{y}$$ My answer was $-\ln(y)+c$ but my professor said it should be $\ln(\frac{c}{y})$ I tried to simplify it more but just end up with $\ln\frac{1}{y}+c$ how did the $C$ became the numerator? 
 A: Both forms are fine. Your professors form is simply obtained from yours by $$\ln \frac 1y+C=\ln \frac 1y+\ln e^C=\ln\frac {e^C}y=\ln\frac{c_1}y$$
Also, $-\ln y=\ln y^{-1}=\ln \frac 1y$
A: Hint: $\log(a)+\log(b)=\log(ab)$
A: we know that $\ln \frac{a}{b}=\ln a-\ln b$
so
$$-\int \frac{dy}{y}=-\ln y +k$$
assume $k=\ln C $ to get
$$-\int \frac{dy}{y}=-\ln y +\ln C=\ln \frac{C}{y}$$
A: The complete answer is 
$$I=-\int \frac{dy}{y}=-\ln |y|+C$$
Inasmuch as we have $-\ln x=\ln (1/x)$, then $I$ can be expressed as 
$$I=\ln \left|\frac{1}{y}\right|+C=\ln \left|\frac{C'}{y}\right|$$
where $\ln |C'|= C$.
The absolute value is important here.  For example, suppose we attempt to evaluate $-\int_{-2}^{-1}\frac{dy}{y}$ as $-\ln(-1)+\ln(-2)$. If we are restricted to real analysis, then these terms don't exist.  However, we know from symmetry that the result should be $\ln 2-\log 1=\ln 2$.  Upon using the appropriate anti-derivative equipped with the absolute values, we obtain
$$-\int_{-2}^{-1}\frac{dy}{y}=\left.\left(-\ln|y|\right)\,\right|_{-2}^{-1}=-\ln|-1|+\ln|-2|=\ln 2$$
as expected!
