I didn't search this video. This video came to me. And I am glad because it helped me understand one of the most thing that is often omitted or neglected.
Took me awhile to understand the reason of putting the minus in front of x. That is because log or ln cant take negative numbers, the only way to take negative numbers is do a double negative. Thx you are actually a unique and fancy teacher compared to Khan, all all other teachers online.
The reason you should write ln|x| is because the derivative of ln|x| is indeed 1/x, and this is defined for all nonzero real x. However, I should also mentioned that the full antiderivative of 1/x REALLY is ln|x| + Csgn(x) + D. This is due to the nonremovable discontinuity at x = 0. Technically, what should have been written in the board, ln(x) + C1 if x > 0, and ln(-x) + C2 if x < 0, and not necessarily assume C1 = C2.
@@mskiptr But this does not address my argument. For the function 1/x specifically, the class of antiderivatives is not the class of functions differing by some value, but rather, the class of functions different by some step function. If I have the step function f(x) = 1 for x > 0, f(x) = -2 for x < 0, then whenever x is defined, f'(x) = 0, so if I define g(x) = ln(|x|) + f(x), then g'(x) = 1/x + f'(x) = 1/x. f is not a constant, but as I just showed you, g still has derivative 1/x.
Well, technically it kind of *is* possible to take the log of a negative number, but it's not necessarily a trivial answer and comes with its own domain restrictions. blackpenredpen demonstrates a proof by one of his subscribers here: th-cam.com/video/k2x7k1WPwXE/w-d-xo.html ln(-x) = ln(x) + iπ(2k+1) ∀ k ∈ Z The log of a negative number just isn't a simple *real* number. It's a complex number. (Which I kinda personally love but hey that's just me.)
And to connect this with the integral of 1/x: The arbitrary constant of integration can be different, on both sides of the asymptote of 1/x, since you cannot integrate across a vertical asymptote, unless the improper integral converges. This means, we get to select an arbitrary constant that nullifies the imaginary part of the complex log. Simply let c = C - i*π*(2*k+1), when x is negative, and let c = C - 2*k*i*π, when x is positive. This allows the integral of 1/x, to be ln(|x|) + C.
![Integrating √[x/(1-x)] by Trigonometric Substitution](http://i.ytimg.com/vi/UEDLMCI-G2c/mqdefault.jpg)
![Integrating √[x/(1-x)] by Trigonometric Substitution](/img/tr.png)
Thank you very much for the great piece of explaination.This is often neglected in most of the textbooks .
I didn't search this video. This video came to me. And I am glad because it helped me understand one of the most thing that is often omitted or neglected.
I was wondering about this for a while and I could never get a straight answer. Thanks for the great explanation
5:01 i like when he says "both" his hands are moving in the ln|x| shape
haha good eye
That's a great explanation of an interesting bit of maths I hadn't properly considered before. Thanks!
Took me awhile to understand the reason of putting the minus in front of x. That is because log or ln cant take negative numbers, the only way to take negative numbers is do a double negative. Thx you are actually a unique and fancy teacher compared to Khan, all all other teachers online.
There's a ratio : x
The reason you should write ln|x| is because the derivative of ln|x| is indeed 1/x, and this is defined for all nonzero real x.
However, I should also mentioned that the full antiderivative of 1/x REALLY is ln|x| + Csgn(x) + D. This is due to the nonremovable discontinuity at x = 0. Technically, what should have been written in the board, ln(x) + C1 if x > 0, and ln(-x) + C2 if x < 0, and not necessarily assume C1 = C2.
Afaik, that +C often stands for all possible constants and the antiderivative is then actually _the whole class of functions_ differing by some value.
@@mskiptr But this does not address my argument. For the function 1/x specifically, the class of antiderivatives is not the class of functions differing by some value, but rather, the class of functions different by some step function. If I have the step function f(x) = 1 for x > 0, f(x) = -2 for x < 0, then whenever x is defined, f'(x) = 0, so if I define g(x) = ln(|x|) + f(x), then g'(x) = 1/x + f'(x) = 1/x. f is not a constant, but as I just showed you, g still has derivative 1/x.
@@angelmendez-rivera351 ah, ok. You're right. I overly focused on the case with two separate functions
brilliant...thank you very much for your great videos...i salute you
It's wonderful when student behaviour is on point and teachers get to TEACH
You are amazing Sir👍
i always thought that its cause 1/x dx can have many different solutions
for example, d/dx(ln(100x)) is just 1/x
hence, 1/x dx can be ln(ax) where a>0
Thank you very much. This is great!
Well, technically it kind of *is* possible to take the log of a negative number, but it's not necessarily a trivial answer and comes with its own domain restrictions. blackpenredpen demonstrates a proof by one of his subscribers here: th-cam.com/video/k2x7k1WPwXE/w-d-xo.html
ln(-x) = ln(x) + iπ(2k+1) ∀ k ∈ Z
The log of a negative number just isn't a simple *real* number. It's a complex number. (Which I kinda personally love but hey that's just me.)
And to connect this with the integral of 1/x:
The arbitrary constant of integration can be different, on both sides of the asymptote of 1/x, since you cannot integrate across a vertical asymptote, unless the improper integral converges. This means, we get to select an arbitrary constant that nullifies the imaginary part of the complex log. Simply let c = C - i*π*(2*k+1), when x is negative, and let c = C - 2*k*i*π, when x is positive.
This allows the integral of 1/x, to be ln(|x|) + C.
great explanation! thank you
great point of view.
Superb sir
Is it possible to understand this if you dont know calculus?
Should be. Imo he used a very simple example for the explanation
Thank you very much.
you are the goat
Thanks Eddie
Thank you
fantastic
awesome
Awesome
grandee chinorris
W mans