I'm reading this book about differential equations, and

>>8594698
Integral of [math] \frac { 1 } { x } = \ln (x) [/math]

Indefinite Integration determines a function up to the addition of an arbitrary constant, since the derivative of a constant is 0.

>>8594698

what is the integral of 1/x dx? what is the integral of k dx where k is a constant?

>>8594709
what is the integral of k dx where k is a constant?

That's what I don't get. I get that the integral of dt is t. That makes sense, but I can't understand what is the integral of dx.

>>8594698
int of (1/x) = ln(x)+c
int of c where is c is a constant = cx+ c
u fuckin dunce get out

>>8594739
But where does the dx go?

>>8594698
Literally integrate both sides.

File: math1.gif (2KB, 213x173px) Image search: [Google]

2KB, 213x173px

>>8594743

I guess I'm saying I don't understand anything about calculus as soon as they start treating dx and dt like variables and moving them around.

I understand what derivatives and integrals are, but Khan mostly just used f(x) and f'(x) notation, and then people suddenly started moving dx and dt around, and I was lost.

>>8594753
Yeah, but in my mind, integrate means that you sandwich the thing between int and dx, so you would get

int 1/x dx dx

What am I not getting here?

>>8594754
I might be wrong, but there is this thing called chain rule, and in the end you can move them around. In your case there is no movement happening.

>>8594754
>I understand what derivatives and integrals are, but Khan mostly just used f(x) and f'(x) notation

The two statements separated by a comma are contradictory.

>>8594762
1/x dx
->
dx / x

It's the notation being used. int f(x) dx. It's like m/s notation -- You can do things like multiply by seconds.

File: Sentido_geometrico_del_diferencial_de_una_funcion.png (20KB, 566x362px) Image search: [Google]

20KB, 566x362px

https://en.wikipedia.org/wiki/Differential_of_a_function

>>8594762
you can integrate something only if it already has a differential. in the OP picture for example he integrated 1/x * dx and -k *dt, and in the pic you're quoting you're integrating 1/x * dx. In other words the dx can't spring from nothing and has to be there because it isn't simply an instruction that you put next to an integral to say 'integrate in this variable', but a mathematical object that you manipulate (as in, for example, multiplying the 1.11 in OP pic by dt on both sides)
you could, for example, look at it geometrically like pic related

>>8594743
u integrated with respect to x, and you integrated with respect to t, theres no point to it anymore

American education, everybody.

>>8594698

If you're not trolling, don't worry about it.
You'll get used to it.

don't worry OP. i got through diff eq's just fine without taking any calc