Primitive de funcții relativ ușoare, însă extraordinare pentru pregătirea de a rezolva altele cu grad de dificultate mai mare. Succes în continuare. Iubesc la infinit matematica, deoarece totul din jurul nostru e matematică.
Yeah, I love this trick! It avoids us to do a long polynomial division D=dq+r ==> D/d = q + r/d D = x^2 d = x^2+1 q = 1 r = -1 D = dq + r ==> x^2 = (x^2+1)*1 - 1 D/d = q + r/d ==> x^2/(x^2+1) = 1 - 1/(x^2+1)
Hola! Sí que entiendo sí! De hecho, cuando creé este canal dudé de si hacerlo en inglés o en español hasta que entendí que no era necesario ningún idioma si escribía el máximo de detalles posibles durante el video 💪 Gracias por tu comentario, espero que disfrutes del canal! Si quieres colaborar con una pequeña ayudita, suscríbete y te lo agradeceré aún más!❤
Hi! It is a little trick that is used when we have polynomials with the same degree in the numerator and in the denominator in order to avoid doing the long polynomial division. The trick consists of completing the numerator with +something-something to get the same numerator and denominator, which is equal to 1, and a new integral where the degree of the denominator is higher than the degree of the numerator. Let me write another example of the trick here: Integral of (x-1)/(x+2) dx = = Integral of (x-1+2-2)/(x+2) dx = = Integral of (x+2-1-2)/(x+2) dx = = Integral of (x+2-3)/(x+2) dx = = Integral of [(x+2)/(x+2) - 3/(x+2)] dx = = Integral of [1 - 3/(x+2)] dx = = Integral of 1 dx - 3*Integral of 1/(x+2) dx = = x - 3*ln|x+2| + C Hope it helped! 💪
Hi Red Black, I am not sure we can do this integral by substitution, the two possible substitutions are u=1+x^2 or x=tan(u) and they make the integral more complicated
Hi Leonrique Pereira! This is the easiest way to do it (when you get use to it :D). We use it to avoid doing a polynomial division D=dq+r: x^2 = 1*(1+x^2) - 1 ==> x^2/(1+x^2) = (1+x^2)/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2) You will find some examples of polynomial division with (1+x^2) in this playlist: ruclips.net/p/PLpfQkODxXi49YODSGByJH2O7dVRgzPKe2
Hola Luis, el 1 lo pongo para señalar que esa función es la que vamos a poner en "dv" cuando se hace la integración por partes. No es necesario escribirlo si no se quiere.
@@IntegralsForYou First i declared u as ln (1+x^2), du= 2x/1+x^2 dx, v = x, DV= dx, then used the formula, the result Was xln (1+x^2) - integral of x.(2x/1+x^2) dx, then i used division to null the x above, in the final i Had the integral of 2 dx, resulting in xln (1+x^2)-2x+C, dont know why this is wrong.
Hi, I don't know the exact mistake you did, but if I am not wrong, you are saying that 2x^2/(1+x^2) = 2 and that's impossible. When you have the integral of 2x^2/(1+x^2) you can do: A. As I do in the video: Integral of 2x^2/(1+x^2) dx = = 2*Integral of x^2/(1+x^2) dx = = 2*Integral of (x^2+1-1)/(1+x^2) dx = = 2*Integral of [ (x^2+1)/(1+x^2) - 1/(1+x^2) ] dx = = 2*Integral of [ 1 - 1/(1+x^2) ] dx = = 2*Integral of 1 dx - Integral of 1/(1+x^2) dx = = 2x - arctan(x) B. Polynomial division: Integral of 2x^2/(1+x^2) dx = = 2*Integral of x^2/(1+x^2) dx = D = dq + r x^2 = (1+x^2)*1 - 1 ==> x^2/(1+x^2) = (1+x^2)*1/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2) I don't know if I helped, but I am sure you did something wrong in your division.
Hola Alejandro! el +1-1 sirve para modificar la integral sin variar el resultado ya que +1-1=0. Esta modificación la hacemos porque la integral de x^2/(1+x^2), que en teoría se resolvería mediante la división de polinomios, hacemos este pequeño truco que se hace en un momento y nos convierte x^2/(1+x^2) = (x^2+1-1)/(x^2+1) = (x^2+1)/(x^2+1) - 1/(x^2+1) = 1 - 1/(1+x^2). Para la integral de x^2/(1+x^2) no sé decirte el resultado a simple vista. Para la integral de 1 - 1/(1+x^2) sí, y es x - arctan(x) Espero haberte ayudado! (y si no, vuelve a preguntar ;-D)
No lo sé, puede que sí, pero se hace tan eterno que no merece la pena saberlo... Se hace largo por culpa de la expresión de dx en función de dt: t = 1+x^2 ==> t-1 = x^2 ==> sqrt(1-t) = x dt = 2x dx ==> dt = 2sqrt(1-t) dx ==> 1/2sqrt(1-t) dt = dx Integral de ln(1+x^2) dx = Integral de ln(t)*1/2sqrt(1-t) dt = ...
Hi Andrew Nicholas! You can try but by the moment I don't know how to continue after 2 steps... : Substitution: x=tan(u) dx= 1/cos^2(u) du Integral of ln(1+x^2) dx = = Integral of ln(1+tan^2(u)) 1/cos^2(u) du = = Integral of ln(1/cos^2(u)) 1/cos^2(u) du = ...
You can use trig sub but the amount of steps are about the same. Though I do think the process is more elegant. x = tan(t) dx = sec^2(t)dt = { ln(1+ tan^2t)*sec^2t dt = { ln(sec^2t)*sec^2t dt = { 2ln(sec t)*sec^2t dt = 2{ ln(sec t)*sec^2t dt u = ln(sec t) du = sec t * tan t / sec t dt = tan t dt dv = sec^2t dt v = tan t = 2(ln(sec t)*tan t - {tan^2t dt) = 2(ln(sec t)*tan t - {(sec^2t - 1)dt) = 2(ln(sec t)*tan t - tan t + t) t = arctan x sec t = √(1 + tan^2t) = √(1 + x^2) = 2ln√(1+x^2)*x - 2x + 2arctanx = 2*1/2 ln(1+x^2)*x - 2(x - arctanx) = x*ln(1+x^2) - 2(x - arctanx) + c
Hi! There is a rule called ILATE that is very useful. This video is a special case, but in general we have two functions multiplying. If you have time, I would recommend try u=first and dv=second and if it doesn't work, then try u=second and dv=first. When you learn it like this, then for the next integrals you will see which is u and dv at first sight. If you are in an exam and you don't have a lot of time, use the ILATE rule 😉
Integral of (1+x^2)sqrt(x) dx = = Integral of (1+x^2)x^(1/2) dx = = Integral of (x^(1/2) + x^(5/2)) dx = = x^(3/2)/(3/2) + x^(7/2)/(7/2) = = (2/3)x^(3/2) + (2/7)x^(7/2) = = (2/21)x^(3/2)(7 + 3x^2) + C Hope it helped! 😊
@@Monsenstar Hi! No need to use substitution, but if still you want to use it it would be: Integral of (1+x^2)sqrt(x) dx = Substitution: u = sqrt(x) ==> u^2 = x ==> u^4 = x^2 du = 1/2sqrt(x) dx = 1/2u dx ==> 2u du = dx = Integral of (1+u^4)u 2u du = = 2*Integral of (u^2 + u^6) du = = 2*(u^3/3 + u^7/7) = = 2*((1/3)u^3 + (1/7)u^7) = = (2/21)*(u^3)(7 + 3u^4) = = (2/21)*((sqrt(x)^3))(7 + 3x^2) = = (2/21)x^(3/2)(7 + 3x^2) + C :-D
Hola NG glz! Si vas a ver las tablas de derivadas, verás que la derivada de arctan(x) es 1/(1+x^2). Por lo tanto, la integral de 1/(1+x^2) es directa y da como solución arctan(x). No sé si esta es la respuesta que buscabas... si no es así, vuélveme a preguntar detallándome un poco más tu duda ;-D
What's the point of integrating by parts if you just make u your original equation anyways? Lol. u should be 1 + x^2 in this case. This is so pointlessly difficult.
Hi! "u" cannot be 1+x^2 because we need two functions multiplying themselves and ln(1+x^2) is not a multiplication. We have the same problem with the integral of ln(x) 👉 ruclips.net/video/a1--E2uOtB8/видео.html . The only way to solve the integral of ln(x) and the integral of ln(1+x^2) by parts is by doing u=ln(1+x^2) and dv = 1dx
Hola Fabriii mussini! En primer lugar hay que entender que sumar y restar la misma cantidad, en este caso 1, no cambia el resultado porque es como si estuvieramos sumando cero. En segundo lugar, hacemos +1-1 porque como en el denominador tenemos 1+x^2 y en el numerador x^2, si le sumamos +1 tendremos lo mismo en el numerador y denominador y se podrá simplificar: (1+x^2)/(1+x^2) = 1 y esto es facil de integrar. Pero si sumamos +1, hay que restarle -1 para que la integral no cambie. Por suerte, 1/(1+x^2) también es facil de integrar. Teoricamente deberiamos hacer la division polinomial D=dq+r, pero si tenemos un poco de ojo antes y podemos utilizar el truco que te acabo de explicar, te ahorras mucho tiempo. Si haces la division polinomial, el resultado seria este: D = x^2 d = 1+x^2 q = 1 r = -1 De modo que D = dq + r ==> x^2 = (1+x^2)*1 - 1 Aunque para integrar usamos D/d = (dq + r)/d = dq/d + r/d = q + r/d ==> x^2/(1+x^2) = 1 - 1/(1+x^2) Para obtener lo mismo más rápidamente usamos el +1-1 : x^2/(1+x^2) = (x^2+1-1)/(1+x^2) = (1+x^2)/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2) Sea cual sea el método que uses, siempre deberías obtener el mismo resultado. Un saludo, espero haberte ayudado! ;-D
Hola Mat! Gracias por el consejo! Ya me lo planteé al principio, pero decidí que prefiero que penséis vosotros por qué hago cada paso y si hay dudas las aclaramos en los comentarios. En mi opinión es una buena técnica para aprender 😊
@@IntegralsForYou Buen punto, solo que seria bueno de apoyo para los que aun no comprenden algunas reglas o que no conocen los métodos de integración por ejemplo.
@@celerinorodriguezgonzalez3508 Sí, cierto! Habrá gente que este método le convenga y otra que no... Al final cada uno tiene su mejor manera de aprender. Agradezco mucho tu aportación, Mat. Un saludo!
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I was stuck in this integral for hours thank you so much for uploading the solving process
My pleasure! Welcome and enjoy the channel! 😊😊
On behalf of all your viewers, thank you. I can't imagine how difficult coming up with some of these solutions.
Thank you very much! It was a little bit hard when starting this channel, but after lots of integrals it become a lot more easier hehe
Primitive de funcții relativ ușoare, însă extraordinare pentru pregătirea de a rezolva altele cu grad de dificultate mai mare. Succes în continuare. Iubesc la infinit matematica, deoarece totul din jurul nostru e matematică.
❤❤❤
Thank you sir! Saved my butt a couple times already 👍
And I am very happy to save it! 😊😜👍
we would appreciate if u did some explaining aswell..
Thank you for save me!!
My pleasure! 💪
Thanks again mate
🤗
ahhh i was struggling trying to evaluate the second integral... the x^2 + 1 - 1 trick really helps!
Yeah, I love this trick! It avoids us to do a long polynomial division D=dq+r ==> D/d = q + r/d
D = x^2
d = x^2+1
q = 1
r = -1
D = dq + r ==> x^2 = (x^2+1)*1 - 1
D/d = q + r/d ==> x^2/(x^2+1) = 1 - 1/(x^2+1)
Thanks man that was helpful
My pleasure! ☺
Estuve buscando videos en español por horas para este ejercicio en específico, no se si entiendes pero gracias
Hola! Sí que entiendo sí! De hecho, cuando creé este canal dudé de si hacerlo en inglés o en español hasta que entendí que no era necesario ningún idioma si escribía el máximo de detalles posibles durante el video 💪 Gracias por tu comentario, espero que disfrutes del canal! Si quieres colaborar con una pequeña ayudita, suscríbete y te lo agradeceré aún más!❤
Gracias, ya me estaba volviendo loco el como resolver esta integrada, Like 1+
Me alegro de que te haya servido de ayuda! 😊Un saludo!
THERE IS A GOD THANK YOU SO MUCH
;-D You're welcome! ;-D
Hi, I have a quick question. Why do you add a "+1 -1" to the x^2 @ 1:47?
Hi! It is a little trick that is used when we have polynomials with the same degree in the numerator and in the denominator in order to avoid doing the long polynomial division.
The trick consists of completing the numerator with +something-something to get the same numerator and denominator, which is equal to 1, and a new integral where the degree of the denominator is higher than the degree of the numerator.
Let me write another example of the trick here:
Integral of (x-1)/(x+2) dx =
= Integral of (x-1+2-2)/(x+2) dx =
= Integral of (x+2-1-2)/(x+2) dx =
= Integral of (x+2-3)/(x+2) dx =
= Integral of [(x+2)/(x+2) - 3/(x+2)] dx =
= Integral of [1 - 3/(x+2)] dx =
= Integral of 1 dx - 3*Integral of 1/(x+2) dx =
= x - 3*ln|x+2| + C
Hope it helped! 💪
Thank you so much!!!
You're welcome! ;-D
Thank you! You're amazing!!!!!!!
You're welcome! Thanks for your comment! 😉
thank u , i would like to know how did u decide what will be dv and v as well
en.wikipedia.org/wiki/Integration_by_parts#LIATE_rule 😉
@IntegralsForYou I use LIPET which is similar log inverse polynomial exp trig so far never steered me wrong
Help me a lot
Happy to know it! 😊😊
lol man love how u write, awesome, keep on doing this amazing job!!!
Thank you :-D
J= xln(1+x^2)-2x +2arctanx +k
Correct! 💪
Crazy notations
Thank you sir ji
My pleasure! 😊
amazing!
Thank you! 😊
Thanks
My pleasure! ❤
How do you integrate it with the substitution method? Can u please make a video or explain? Thanks
Hi Red Black, I am not sure we can do this integral by substitution, the two possible substitutions are u=1+x^2 or x=tan(u) and they make the integral more complicated
1:45 awesome
😊Thanks! 😊
GOOOOD 👍
Thaaaaanks! 😉😜
Appreciate your video, but i wonder why some hold their pen so weirdly. 🤔
Hehe thank you! Honestly, I don't know why I hold it that way... 😅
Gracias!
Un placer! :-D
Is that thing about +1-1 a regular and common way to solve it, or there are easier ways?
Hi Leonrique Pereira! This is the easiest way to do it (when you get use to it :D). We use it to avoid doing a polynomial division D=dq+r:
x^2 = 1*(1+x^2) - 1 ==> x^2/(1+x^2) = (1+x^2)/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2)
You will find some examples of polynomial division with (1+x^2) in this playlist: ruclips.net/p/PLpfQkODxXi49YODSGByJH2O7dVRgzPKe2
thank u, save my life. ^^
my hero
Thanks! 😊
dudeeeeeee wtf? where is your voice?it would be a great video if you just could explain a little bit. Add value to your videos...
honestly , It's totally understable without his voice ..
De donde salio el numero uno que pones despues de l resultado primero? Responde plis
Hola Luis, el 1 lo pongo para señalar que esa función es la que vamos a poner en "dv" cuando se hace la integración por partes. No es necesario escribirlo si no se quiere.
@@IntegralsForYou gracias gracias muchas gracias enserio...entrego trabajos en media hora y me salvaste la vida bro
LUIS GONZALEZ Me alegro mucho haberte servido de ayuda! Ya me dirás qué nota obtienes! 😉
I dont get it, why the resolution isnt xln (1+x^2)-2x+k? What am i doing wrong?
Hi Xdetonando! Could you please write the steps you did? Thanks!
@@IntegralsForYou First i declared u as ln (1+x^2), du= 2x/1+x^2 dx, v = x, DV= dx, then used the formula, the result Was xln (1+x^2) - integral of x.(2x/1+x^2) dx, then i used division to null the x above, in the final i Had the integral of 2 dx, resulting in xln (1+x^2)-2x+C, dont know why this is wrong.
Hi, I don't know the exact mistake you did, but if I am not wrong, you are saying that 2x^2/(1+x^2) = 2 and that's impossible. When you have the integral of 2x^2/(1+x^2) you can do:
A. As I do in the video:
Integral of 2x^2/(1+x^2) dx =
= 2*Integral of x^2/(1+x^2) dx =
= 2*Integral of (x^2+1-1)/(1+x^2) dx =
= 2*Integral of [ (x^2+1)/(1+x^2) - 1/(1+x^2) ] dx =
= 2*Integral of [ 1 - 1/(1+x^2) ] dx =
= 2*Integral of 1 dx - Integral of 1/(1+x^2) dx =
= 2x - arctan(x)
B. Polynomial division:
Integral of 2x^2/(1+x^2) dx =
= 2*Integral of x^2/(1+x^2) dx =
D = dq + r
x^2 = (1+x^2)*1 - 1
==> x^2/(1+x^2) = (1+x^2)*1/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2)
I don't know if I helped, but I am sure you did something wrong in your division.
great!
;-D
profe buen video solo que no entendi el por que pone (+1-1) gracias me esta ayudando :3
Hola Alejandro! el +1-1 sirve para modificar la integral sin variar el resultado ya que +1-1=0. Esta modificación la hacemos porque la integral de x^2/(1+x^2), que en teoría se resolvería mediante la división de polinomios, hacemos este pequeño truco que se hace en un momento y nos convierte x^2/(1+x^2) = (x^2+1-1)/(x^2+1) = (x^2+1)/(x^2+1) - 1/(x^2+1) = 1 - 1/(1+x^2).
Para la integral de x^2/(1+x^2) no sé decirte el resultado a simple vista.
Para la integral de 1 - 1/(1+x^2) sí, y es x - arctan(x)
Espero haberte ayudado! (y si no, vuelve a preguntar ;-D)
Y no se puede hacer esta integral cambiando x^2 + 1 = t ??
No lo sé, puede que sí, pero se hace tan eterno que no merece la pena saberlo... Se hace largo por culpa de la expresión de dx en función de dt:
t = 1+x^2 ==> t-1 = x^2 ==> sqrt(1-t) = x
dt = 2x dx ==> dt = 2sqrt(1-t) dx ==> 1/2sqrt(1-t) dt = dx
Integral de ln(1+x^2) dx = Integral de ln(t)*1/2sqrt(1-t) dt = ...
can you use trig sub ?
Hi Andrew Nicholas! You can try but by the moment I don't know how to continue after 2 steps... :
Substitution:
x=tan(u)
dx= 1/cos^2(u) du
Integral of ln(1+x^2) dx =
= Integral of ln(1+tan^2(u)) 1/cos^2(u) du =
= Integral of ln(1/cos^2(u)) 1/cos^2(u) du = ...
You can use trig sub but the amount of steps are about the same. Though I do think the process is more elegant.
x = tan(t)
dx = sec^2(t)dt
= { ln(1+ tan^2t)*sec^2t dt
= { ln(sec^2t)*sec^2t dt
= { 2ln(sec t)*sec^2t dt
= 2{ ln(sec t)*sec^2t dt
u = ln(sec t)
du = sec t * tan t / sec t dt = tan t dt
dv = sec^2t dt
v = tan t
= 2(ln(sec t)*tan t - {tan^2t dt)
= 2(ln(sec t)*tan t - {(sec^2t - 1)dt)
= 2(ln(sec t)*tan t - tan t + t)
t = arctan x
sec t = √(1 + tan^2t) = √(1 + x^2)
= 2ln√(1+x^2)*x - 2x + 2arctanx
= 2*1/2 ln(1+x^2)*x - 2(x - arctanx)
= x*ln(1+x^2) - 2(x - arctanx) + c
how can i understand what i have to consider a s U or V
Hi! There is a rule called ILATE that is very useful. This video is a special case, but in general we have two functions multiplying. If you have time, I would recommend try u=first and dv=second and if it doesn't work, then try u=second and dv=first. When you learn it like this, then for the next integrals you will see which is u and dv at first sight. If you are in an exam and you don't have a lot of time, use the ILATE rule 😉
Integral (1+x²)√x dx
how
Integral of (1+x^2)sqrt(x) dx =
= Integral of (1+x^2)x^(1/2) dx =
= Integral of (x^(1/2) + x^(5/2)) dx =
= x^(3/2)/(3/2) + x^(7/2)/(7/2) =
= (2/3)x^(3/2) + (2/7)x^(7/2) =
= (2/21)x^(3/2)(7 + 3x^2) + C
Hope it helped! 😊
@@IntegralsForYou thank u, u are the kind human exist. But what if we use du dx ?
@@Monsenstar Hi! No need to use substitution, but if still you want to use it it would be:
Integral of (1+x^2)sqrt(x) dx =
Substitution:
u = sqrt(x) ==> u^2 = x ==> u^4 = x^2
du = 1/2sqrt(x) dx = 1/2u dx ==> 2u du = dx
= Integral of (1+u^4)u 2u du =
= 2*Integral of (u^2 + u^6) du =
= 2*(u^3/3 + u^7/7) =
= 2*((1/3)u^3 + (1/7)u^7) =
= (2/21)*(u^3)(7 + 3u^4) =
= (2/21)*((sqrt(x)^3))(7 + 3x^2) =
= (2/21)x^(3/2)(7 + 3x^2) + C
:-D
god!
;-D
¿Porque en el resultado sale arctan(x) ?
Hola NG glz! Si vas a ver las tablas de derivadas, verás que la derivada de arctan(x) es 1/(1+x^2). Por lo tanto, la integral de 1/(1+x^2) es directa y da como solución arctan(x). No sé si esta es la respuesta que buscabas... si no es así, vuélveme a preguntar detallándome un poco más tu duda ;-D
What's the point of integrating by parts if you just make u your original equation anyways? Lol. u should be 1 + x^2 in this case. This is so pointlessly difficult.
Hi! "u" cannot be 1+x^2 because we need two functions multiplying themselves and ln(1+x^2) is not a multiplication. We have the same problem with the integral of ln(x) 👉 ruclips.net/video/a1--E2uOtB8/видео.html . The only way to solve the integral of ln(x) and the integral of ln(1+x^2) by parts is by doing u=ln(1+x^2) and dv = 1dx
From which country you belong😀?
Hi! I'm from Spain 😉
@@IntegralsForYou Lovely country♥️
@Shreya Verma 😊
alguien me puede decir de donde sale ese +1 -1??
Hola Fabriii mussini! En primer lugar hay que entender que sumar y restar la misma cantidad, en este caso 1, no cambia el resultado porque es como si estuvieramos sumando cero. En segundo lugar, hacemos +1-1 porque como en el denominador tenemos 1+x^2 y en el numerador x^2, si le sumamos +1 tendremos lo mismo en el numerador y denominador y se podrá simplificar: (1+x^2)/(1+x^2) = 1 y esto es facil de integrar. Pero si sumamos +1, hay que restarle -1 para que la integral no cambie. Por suerte, 1/(1+x^2) también es facil de integrar.
Teoricamente deberiamos hacer la division polinomial D=dq+r, pero si tenemos un poco de ojo antes y podemos utilizar el truco que te acabo de explicar, te ahorras mucho tiempo. Si haces la division polinomial, el resultado seria este:
D = x^2
d = 1+x^2
q = 1
r = -1
De modo que D = dq + r ==> x^2 = (1+x^2)*1 - 1
Aunque para integrar usamos D/d = (dq + r)/d = dq/d + r/d = q + r/d ==> x^2/(1+x^2) = 1 - 1/(1+x^2)
Para obtener lo mismo más rápidamente usamos el +1-1 :
x^2/(1+x^2) = (x^2+1-1)/(1+x^2) = (1+x^2)/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2)
Sea cual sea el método que uses, siempre deberías obtener el mismo resultado.
Un saludo, espero haberte ayudado! ;-D
no entendi el resultado :(
Si me dices que no entiendes exactamente intentaré echarte una mano :-D
heloo i'm looking for this x/(x^2+x+1) integral
Hi Fatima!
Integral of x/(x^2+x+1) dx =
= (1/2)Integral of (2x)/(x^2+x+1) dx =
= (1/2)Integral of (2x+1-1)/(x^2+x+1) dx =
= (1/2)[ Integral of (2x+1)/(x^2+x+1) dx - Integral of 1/(x^2+x+1) dx ] =
= (1/2)[ ln|x^2+x+1| - ruclips.net/video/hoaCmqkp094/видео.html ] =
= (1/2)[ ln|x^2+x+1| - (2/sqrt(3))arctan((2x+1)/sqrt(3)) ] =
= (1/2)ln|x^2+x+1| - (1/sqrt(3))arctan((2x+1)/sqrt(3)) + C
:-D
thank u so much
You're welcome!
спас брат
😀
consejo agregar explicacion oral
Hola Mat! Gracias por el consejo! Ya me lo planteé al principio, pero decidí que prefiero que penséis vosotros por qué hago cada paso y si hay dudas las aclaramos en los comentarios. En mi opinión es una buena técnica para aprender 😊
@@IntegralsForYou Buen punto, solo que seria bueno de apoyo para los que aun no comprenden algunas reglas o que no conocen los métodos de integración por ejemplo.
@@celerinorodriguezgonzalez3508 Sí, cierto! Habrá gente que este método le convenga y otra que no... Al final cada uno tiene su mejor manera de aprender. Agradezco mucho tu aportación, Mat. Un saludo!
So easy....its child's play
just like shitting in the toilet, but your people dont seem to do it that much, pajeet
махталитет )
It's wrong if you're integration is between infinite
Hi! In infinite the integral diverges, yes
WHAT THE FUCK!!
A positive WTF or a negative WTF? 😜
thank you you saw greath
You're welcome! :-D
l
O
smurf.... why smurfing
The worst way to do this ever
I didn't know there are other ways to do it...! Can you share them, please?