Hola! SI haces dv=(arctan(x))^2 dx tienes que integrar (arctan(x))^2 para obtener v, y siguiendo por ese camino se complica más en comparación al caso dv = x*dx
@@IntegralsForYou Sir If you please i have a problem with Decomposing rational fraction When the power of X is big Can you show me a way not to waste time while solving it Or another method Please and thank you !
Here we have to use partial fraction decomposition method. It is true that in this case it is a bit long but it is not a waste of time, since it is the only method we can use here: (x+1)/[(x^2+1)(x-1)^4] = A/(x-1) + B/(x-1)^2 + C/(x-1)^3 + D/(x-1)^4 + (Ex+F)/(x^2+1) = ...
Hola Enrique! Como el la resolución se estaba alargando mucho y me estaba quedando sin espacio tuve que hacerla directamente... Teóricamente se hace mediante sustitución u = arctan(x). Te dejo aquí el enlace del video: Integral de arctan(x)/(1+x^2) dx = ua-cam.com/video/11yRV--qkso/v-deo.html Un saludo y gracias por preguntar!
Hello I had the same question as Enrique, but I think we "just" have to see that we already did it for arctan(x)² and we obtain the same thing times 2 so we end up dividing it by 2.
Hola Carlos! Este canal fue creado para tener una lista de integrales y sus soluciones para gente que necesite practicar las técnicas y métodos que ya ha aprendido. Si esta persona no sabe por qué hago un paso, le invito a que reflexione lo máximo que pueda y si no consigue saber por qué, que lo pregunte en modo comentario. En otras palabras, no lo explico para no ponerlo tan fácil :-D. Si necesitas algo no dudes en preguntar. Un saludo y gracias por ver y comentar el video!
Hi Shubhamm Rawatt! Could you please write the expression with parentheses? I also don't understand if there is something between the 1 and √ ... Thanks!
Sorry again, I don't understand what does "into" mean in that case. You mean (x-1)/(x+1sqrt(x^3 + x^2 + x)) or (x-1)/(x+1/sqrt(x^3 + x^2 + x)) or something different? Sorry...
Sorry I don't give my number in this channel, you can contact me like we are doing, commenting on the video. However, any of the integrals written before have solution in terms of standard functions (you can check it on the WolframAlpha integrator online). If I understand well, you mean the integral of (x-1)/( (x+1)sqrt(x^3 + x^2 + x) ). As I have already said, there isn't a solution for that one.
Muchas gracias, tenía problemas con ese ejercicio.
De nada, un placer!
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Спасибо, индус
He is not indian
excelente video, este ejercicio no me salía, buena página, éxitos
Muchas gracias Alexander, se agradecen mucho los comentarios como el tuyo ;-D
THANKSS VERY MUCH!
You're welcome, BamPintor! ;-D
excelente video :D
Gracias Julio!
merci
De rien! 😎
GREATTTTTTTT
Thanksssss! 💪😊
sir plz explain, is it possible to find integration of sqrt(arctan(x))
Hi! I don't think it is possible...
Thanks
You're welcome! ;-D
pero u no deveria ser x y v arcotangente?
Hola! SI haces dv=(arctan(x))^2 dx tienes que integrar (arctan(x))^2 para obtener v, y siguiendo por ese camino se complica más en comparación al caso dv = x*dx
yyay i did it like you ;-)
Yeah! ;-D 😉
@@IntegralsForYou Sir If you please i have a problem with Decomposing rational fraction When the power of X is big Can you show me a way not to waste time while solving it Or another method Please and thank you !
Do you have any example?
@@IntegralsForYou yess here it is :
(x+1)/[ (x-1)⁴ * (x²+1) ]
Here we have to use partial fraction decomposition method. It is true that in this case it is a bit long but it is not a waste of time, since it is the only method we can use here:
(x+1)/[(x^2+1)(x-1)^4] = A/(x-1) + B/(x-1)^2 + C/(x-1)^3 + D/(x-1)^4 + (Ex+F)/(x^2+1) = ...
Tengo una duda en el minuto 6:00 , quisiera saber como realizaste la integral de Arctan(x)*1/1+x^2 dx , porfavor
Hola Enrique!
Como el la resolución se estaba alargando mucho y me estaba quedando sin espacio tuve que hacerla directamente... Teóricamente se hace mediante sustitución u = arctan(x). Te dejo aquí el enlace del video:
Integral de arctan(x)/(1+x^2) dx = ua-cam.com/video/11yRV--qkso/v-deo.html
Un saludo y gracias por preguntar!
Hello I had the same question as Enrique, but I think we "just" have to see that we already did it for arctan(x)² and we obtain the same thing times 2 so we end up dividing it by 2.
x que no explicas lo que haces :-)
Hola Carlos! Este canal fue creado para tener una lista de integrales y sus soluciones para gente que necesite practicar las técnicas y métodos que ya ha aprendido. Si esta persona no sabe por qué hago un paso, le invito a que reflexione lo máximo que pueda y si no consigue saber por qué, que lo pregunte en modo comentario. En otras palabras, no lo explico para no ponerlo tan fácil :-D. Si necesitas algo no dudes en preguntar. Un saludo y gracias por ver y comentar el video!
me parece perfecto
plz integrate this x-1/x+1√x³+x²+x
Hi Shubhamm Rawatt! Could you please write the expression with parentheses? I also don't understand if there is something between the 1 and √ ... Thanks!
sure😀
integration of( x-1) upon( x+1into root x³+x²+x)
Sorry again, I don't understand what does "into" mean in that case. You mean (x-1)/(x+1sqrt(x^3 + x^2 + x)) or (x-1)/(x+1/sqrt(x^3 + x^2 + x)) or something different? Sorry...
into means multiply
can u give me ur WhatsApp no.
Sorry I don't give my number in this channel, you can contact me like we are doing, commenting on the video. However, any of the integrals written before have solution in terms of standard functions (you can check it on the WolframAlpha integrator online).
If I understand well, you mean the integral of (x-1)/( (x+1)sqrt(x^3 + x^2 + x) ). As I have already said, there isn't a solution for that one.
crack
:-D
Thanks
You're welcome! 😊