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?
 A: I like to think of it as cancelling out the $dx$ but what is actually happening is this:
$$I=\int\frac{dy}{dx}\frac{1}{y^2}dx$$
$u=y(x)\Rightarrow du=y'dx\therefore dx=\frac{du}{y'}$
$$I=\int\frac{y'}{y'}\frac{1}{u^2}du=\int\frac{1}{u^2}du=-\frac{1}{u}+C=-\frac 1y+C$$
A: 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
$$g'(x)=-x^{-2}=-\frac{1}{x^2}$$
A: No. You shouldn't just cancel out the $\text{d}x$, since they aren't numbers, although they can be defined so it's possible to do such cancellation. But for now you should just subsitute.
On the other hand. you MUST NOT cancel out $\text{d}y(x)$ with $y(x)$. That's completely wrong, since $\text{d}$ is an operator, not something you multiply. .They are two different things.
That said, what you should do to make clearer those kind of integrals (and avoid getting confused) is to subsitute as I said earlier. Take $u(x)=y(x)$. Then $u'=\dfrac{\text{d}u(x)}{\text{d}x}=\dfrac{\text{d}y(x)}{\text{d}x}$. So
$\int \frac{\frac{d y(x)}{dx}}{y(x)^2}dx = \int \frac{u'}{u^2}\text{d}x=-\dfrac{1}{u}+C=-\frac{1}{y(x)}+C$, where the last step is ,ade undoing the substitution, and $C$ is a real constant. If you set $C=0$ you have a particular antiderivative for $\dfrac{\frac{y(x)}{\text{d}x}}{y(x)^2}$.
A: Cancel differential $dx$ in numerator and denominator in the finite differential sense,before returning to infinitesimals.
$$\int \dfrac{d \; y(x)}{y(x)^2} $$
If $y(x)=v, y'(x)=v'$ then directly integrating
$$\dfrac{-1}{v}+c$$
$$= -\dfrac{1}{y(x)}+c$$
Remember $ \int x^n dx= x^{n+1}/(n+1)+c$
