How is $\int \frac{\frac{d y(x)}{dx}}{y(x)^2}dx = -\frac{1}{y(x)}$ I just saw wolfram-alpha making following statement...
$\int \frac{\frac{d y(x)}{dx}}{y(x)^2}dx = -\frac{1}{y(x)}$
But I am very unsure how to get there.
I always thought I can cancel out the $dx$ and call it a day. Like so:
$\int \frac{\frac{d y(x)}{dx}}{y(x)^2}dx =  \int \frac{d y(x)}{y(x)^2}=\int \frac{d 1}{y(x)}=\frac{1}{y(x)}$
I suppose that's wrong, so where does the minus come from?

You make a mistake when you think that $dy(x)$ is a product of $ d $ by $ y(x)$ and simplify.
You want to find
$$J=\int \frac{y'(x)}{y^2(x)}dx$$
So, you will make the substitution
$$t=y(x)$$
with
$$dt=y'(x)dx$$
It becomes
$$J=\int \frac{dt}{t^2}=-\frac 1t+C$$
$$=-\frac{1}{y(x)}+C$$
The minus comes from the fact that the derivative of $ g(x)=\frac{1}{x}=x^{-1} $ is
$$g'(x)=-x^{-2}=-\frac{1}{x^2}$$
