Hi! Here you have the solution in three different ways: 1. Integral of 1/(1-x) dx = = - Integral of -1/(1-x) dx = = -ln|1-x| + C 2. Integral of 1/(1-x) dx = = Integral of -1/(x-1) dx = = - Integral of 1/(x-1) dx = = - integralsforyou.com/integrals?v=N375VoZod6A = = -ln|x-1| + C 3. Integral of 1/(1-x) dx = Substitution: u = 1-x du = -dx ==> -du = dx = Integral of 1/u (-du) = = - Integral of 1/u du = = -ln|u| = = -ln|1-x| + C Hope it helped! ;-)
Hi! Because when we do u-substitution we have to derive in both sides. Let's imagine we have the next substitution: f(u) = g(x) In order to get the "du" and "dx" we derive in both sides and we add the "du" and "dx" on their side: f'(u)du = g'(x)dx (where f'(u) is the derivative of f(u) and g'(x) is the derivative of g(x)) In our case in the video we have f(u)=u (where f'(u)=1) and g(x)=x-1 (where g'(x)=1-0). Then: 1*du = (1-0)*dx ==> du = dx Hope it helped! ;-D
@@mer2760 On the left we derive with respect to "u" and on the right we derive with respect to "x". For example, if we do "sin(u) = x^2" then "cos(u) du = 2x dx". Keep in mind that the most used substitutions are the u-substitution and the trig substitution, then you will mostly find substitutions like "u=f(x)" or "x=g(u)". Again, for example: u-substitution: u=x^2 ==> du = 2x dx trig substitution: x=sin(u) ==> dx = cos(u)du
Hi! If it was -ln(x-1) then its derivative would be: Derivative of -ln(x-1) = = - Derivative of ln(x-1) = = - 1/(x-1) When we derive ln(x-1) we must multiply by the derivative of x-1 but since it is equal to 1 (and not -1) the derivative of ln(x-1) is 1/(x-1). Hope it helped! 💪
@@arindamn4880 Oh, nice! Glad we found the issue! 💪 I did the video for the integral of 1/(1-x) here th-cam.com/video/WJoxybUK_6s/w-d-xo.html , if you are interested! 🙂
Hi! Here you have the solution: Integral of 1/(1-x) dx = Substitution: u = 1-x du = -dx ==> -du = dx = Integral of 1/u (-du) = = - Integral of 1/u du = = - ln|u| = = - ln|1-x| + C ;-D
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U never know how much u helped me ☺️☺️
hehe my pleasure! Enjoy the channel! 💪
thnk buddy these types of shorts are really helpful . keep going
Thanks! I'll keep posting! Enjoy the channel! 😊
Como diríamos en Argentina, "cortito y al pie".
Gracias!
jajaj esa era la idea cuando pensé en este canal, directo al grano jeje Un saludo!!
Tqsm👍
You're welcome! 😊
😊thanks
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Thanks
You're welcome, Mohameed! 👍👍
what is the integration of 1/(1-x) ?
Hi! Here you have the solution in three different ways:
1.
Integral of 1/(1-x) dx =
= - Integral of -1/(1-x) dx =
= -ln|1-x| + C
2.
Integral of 1/(1-x) dx =
= Integral of -1/(x-1) dx =
= - Integral of 1/(x-1) dx =
= - integralsforyou.com/integrals?v=N375VoZod6A =
= -ln|x-1| + C
3.
Integral of 1/(1-x) dx =
Substitution:
u = 1-x
du = -dx ==> -du = dx
= Integral of 1/u (-du) =
= - Integral of 1/u du =
= -ln|u| =
= -ln|1-x| + C
Hope it helped! ;-)
@@IntegralsForYou thnx :)
@PT064 My pleasure! 😎
Sir, i wanted to ask that, can we write x-1 as (√x²)-1²and then apply formula of x²-a² to solve the question ❓
Hi, I'm afraid you cannot... Well, you can but it won't improve the expression or the way to solve it...
@@IntegralsForYou Thank you sir, for taking the time to explain this.
@Priyanshu Trivedi My plesure! 😊
Thanks a lot!
You're welcome! 😉
Why dm=dx
Hi! Because when we do u-substitution we have to derive in both sides. Let's imagine we have the next substitution: f(u) = g(x)
In order to get the "du" and "dx" we derive in both sides and we add the "du" and "dx" on their side:
f'(u)du = g'(x)dx (where f'(u) is the derivative of f(u) and g'(x) is the derivative of g(x))
In our case in the video we have f(u)=u (where f'(u)=1) and g(x)=x-1 (where g'(x)=1-0). Then:
1*du = (1-0)*dx ==> du = dx
Hope it helped! ;-D
@@IntegralsForYou when we derive, we derive with respect to what?
@@mer2760 On the left we derive with respect to "u" and on the right we derive with respect to "x". For example, if we do "sin(u) = x^2" then "cos(u) du = 2x dx".
Keep in mind that the most used substitutions are the u-substitution and the trig substitution, then you will mostly find substitutions like "u=f(x)" or "x=g(u)". Again, for example:
u-substitution: u=x^2 ==> du = 2x dx
trig substitution: x=sin(u) ==> dx = cos(u)du
@@IntegralsForYou i get it now thank you
is not it (-ln)?
Hi! If it was -ln(x-1) then its derivative would be:
Derivative of -ln(x-1) =
= - Derivative of ln(x-1) =
= - 1/(x-1)
When we derive ln(x-1) we must multiply by the derivative of x-1 but since it is equal to 1 (and not -1) the derivative of ln(x-1) is 1/(x-1).
Hope it helped! 💪
It was 1/1-x
@@IntegralsForYou thank you teacher, I understand now
@@arindamn4880 Oh, nice! Glad we found the issue! 💪
I did the video for the integral of 1/(1-x) here th-cam.com/video/WJoxybUK_6s/w-d-xo.html , if you are interested! 🙂
@@IntegralsForYou I got it now, thanks teacher.
thanks a lot
My pleasure! 💪💪
why is the Nat. Log included ??...
Hi! Because since the derivative of 1/u is ln(u) then the integral of 1/u is ln(u) 😉
Being a lefty it is hard seeing people writing with right hand
You can rotate the screen 😜😜
5/3/22
149am
DEUS, VOCÊ FAZ OQ A PRIME FAZ !!
Lo siento, no lo entiendo, no hablo portugués...
Write a little bigger that makes it more clear
Hi! I'm sorry, I'll try!
@@IntegralsForYou yup mate😍
@Akshaj Dabral 😉
Nasty hand writing sir
Okey, thank you for the information! 😉
Help! Please : |dx/1-x
Hi! Here you have the solution:
Integral of 1/(1-x) dx =
Substitution:
u = 1-x
du = -dx ==> -du = dx
= Integral of 1/u (-du) =
= - Integral of 1/u du =
= - ln|u| =
= - ln|1-x| + C
;-D
@@IntegralsForYou Gracias Amigo! 🙏
@Josue Pantoja De nada, un placer! 😊
Bakwas videos
Thank you so much!
My pleasure! 😊
Thanks
My pleasure! 😊