Once you have this anti derivative, then through a very easy u-sub you can then obtain the anti derivative for 1/(x^3-a^3) in terms of "a". Saves a lot of work.
I tried reading through the comments to see if you already answered these questions, but I didn't see it asked. -@05:47 following the (*), how did you get a (1/6) coefficient and (2x+4) numerator? -How did you know to split the 4 to (1+3)? -How was I suppose to know to work with (x+(1/2))^2? Was it experience in re-writing denominators like (x^2+x+1)? Also just want to thank you creating this UA-cam channel! It's very direct, and it's been helping me through my Cal II class.
Hi! First of all, thank you for the last part of your comment, it really motivates me to keep this channel growing. Ok, let's see: From 5:47 we have (1/3)*Integral of (x+2)/(x^2+x+1) and we want to calculate it. The first thing we want is to have the derivative of x^2+x+1 on the numerator and if we are lucky, just by multiplying inside and outside the integral we can get the derivative on the numerator. f(x)=x^2+x+1 ==> f'(x)=2x+1 . Since we have "x" and not "2x", we multiply/divide inside/outside the integral by 2: (1/3)*Integral of (x+2)/(x^2+x+1) dx = = (1/3)*(1/2)*Integral of (2x+4)/(x^2+x+1) dx = ... Now we have 2x+1 instead of 2x+4. Then, we can do 2x+4=2x+1+3 and split the 3 into another fraction: = (1/6)*Integral of (2x+4)/(x^2+x+1) dx = = (1/6)*Integral of (2x+1+3)/(x^2+x+1) dx = = (1/6)*[Integral of (2x+1)/(x^2+x+1) dx + 3*Integral of 1/(x^2+x+1) dx ] = ... Now we have 1/(x^2+x+1) which obviously does not have 2x+1 on the numerator and it will be impossible to get it. Then we need to write x^2+x+1 as 1+(something)^2 in order to use the arctan formula: x^2+x+1 a^2+2ab+b^2 ==> a^2 = x^2 ==> a = x 2ab = x ==> 2xb = x ==> 2b = 1 ==> b = 1/2 Let's see what happens if we do (x+1/2)^2: (x+1/2)^2 = x^2 + x + 1/4 =====then====> 3/4 + (x+1/2)^2 = x^2 + x + 1 I did the video of 1/(x^2+x+1) here if you want to take a look ;) ua-cam.com/video/hoaCmqkp094/v-deo.html Hope it helped! ;-)
Hi! In the video I use the Ruffini's rule which is used when we want to divide a polynomial by x-a (where "a" is a constant): en.wikipedia.org/wiki/Ruffini's_rule
Hi! I would say practice a lot asking and answering by yourself why we are doing each step. At the end you will see that we don't use a lot of tricks to solve integrals...
Hi Lindsey Wigand! I don't know if I understand your question but I'll explain this step with more detail: We need something like 1/(1+(sth)^2) and we have this 4/3 multiplying, so we need to put it inside the square. So: (1/2)Integral of 1/[ (3/4)(1+(4/3)(x + 1/2)^2 ) ] dx = = (1/2)Integral of 4/[ 3(1+(4/3)(x + 1/2)^2 ) ] dx = = (1/2)(4/3)Integral of 1/[ 1+(4/3)(x + 1/2)^2 ] dx = = (2/3)Integral of 1/[ 1+ ((2/sqrt(3))^2)(x + 1/2)^2 ] dx = = (2/3)Integral of 1/[ 1+ ( (2/sqrt(3))(x + 1/2) )^2 ] dx = = (2/3)Integral of 1/[ 1+ ( (2/sqrt(3))x + (2/sqrt(3))(1/2) )^2 ] dx = = (2/3)Integral of 1/[ 1+ ( (2/sqrt(3))x + 1/sqrt(3) )^2 ] dx = ... I did something like: (a/b)(c+d)^2 = = ((sqrt(a)/sqrt(b))^2)(c+d)^2 = = [ (sqrt(a)/sqrt(b))(c+d) ]^2 = = [ (sqrt(a)/sqrt(b))c + (sqrt(a)/sqrt(b))d ]^2 Hope it helped... and if not, could you please ask me again with more details about your question? Thanks! ;-D
Hi, I know that it would be easier for you if I talked but I prefer to let you think why am I doing each step and if you don't understand it you can ask me in a comment. In my opinion, it is the best way to practice integration 😊
Por supuesto! Aunque quizás yo aprovecharía e intentaría incluir el 3/4 en la sustitución con "u": Integral de (x+2)/[ (x+1/2)^2 + 3/4 ] dx = = Integral de (x+2)/(3/4)*[ (4/3)(x+1/2)^2 + 1 ] dx = = (4/3)*Integral de (x+2)/[ ( (2/sqrt(3))(x+1/2))^2 + 1 ] dx = = (4/3)*Integral de (x+2)/[ ( (2/sqrt(3))x + 1/sqrt(3) )^2 + 1 ] dx = Sustitución: u = (2/sqrt(3))x + 1/sqrt(3) = (2x+1)/sqrt(3) ==> (sqrt(3)u-1)/2 = x du = (2/sqrt(3))dx ==> (sqrt(3)/2)du = dx = (4/3)*Integral de ((sqrt(3)u-1)/2 + 2)/(u^2 + 1) (sqrt(3)/2)du = = (4/3)*(sqrt(3)/2)*Integral de ( (sqrt(3)/2)u - 1/2 + 2)/(u^2 + 1) du = = (2/sqrt(3))*Integral de ( (sqrt(3)/2)u + 3/2)/(u^2 + 1) du = ... Espero haberte ayudado! Un saludo!
@@IntegralsForYou Sí, muchas gracias por darte el tiempo de responder. Si no es mucha molestia, el resultado de la integral, es - ln| 1 / sqrt(u^2 +1 ) | + sqrt(3) arctan(u) y realizando la sustitución de u, entonces es -ln| 1 / [sqrt( (2/sqrt(3) x + 1/sqrt(3) )^2 +1 )] | + arctan(2/sqrt(3) x + 1/sqrt(3) )^2 , cierto?
@@joseorduna7010 Hola! Hoy tengo algo más de tiempo para poder completar mi respuesta anterior: Integral de (x+2)/[ (x+1/2)^2 + 3/4 ] dx = = Integral de (x+2)/(3/4)*[ (4/3)(x+1/2)^2 + 1 ] dx = = (4/3)*Integral de (x+2)/[ ( (2/sqrt(3))(x+1/2))^2 + 1 ] dx = = (4/3)*Integral de (x+2)/[ ( (2/sqrt(3))x + 1/sqrt(3) )^2 + 1 ] dx = Sustitución: u = (2/sqrt(3))x + 1/sqrt(3) = (2x+1)/sqrt(3) ==> (sqrt(3)u-1)/2 = x du = (2/sqrt(3))dx ==> (sqrt(3)/2)du = dx = (4/3)*Integral de ((sqrt(3)u-1)/2 + 2)/(u^2 + 1) (sqrt(3)/2)du = = (4/3)*(sqrt(3)/2)*Integral de ( (sqrt(3)/2)u - 1/2 + 2)/(u^2 + 1) du = = (2/sqrt(3))*Integral de ( (sqrt(3)/2)u + 3/2)/(u^2 + 1) du = = (2/sqrt(3))*[ (sqrt(3)/2)*Integral de u/(u^2+1) du + (3/2)*Integral de 1/(u^2 + 1) du ] = = Integral de u/(u^2+1) du + sqrt(3)*Integral de 1/(u^2+1) du = = (1/2)*Integral de 2u/(u^2+1) du + sqrt(3)*Integral de 1/(u^2+1) du = = (1/2)*ln|u^2+1| + sqrt(3)*arctan(u) = = (1/2)*ln| [(2/sqrt(3))x + 1/sqrt(3)]^2 + 1| + sqrt(3)*arctan((2/sqrt(3))x + 1/sqrt(3)) + C = Hasta aquí podríamos decir que ya hemos acabado, pero simplifiquemos un poco, no? ;-) = (1/2)*ln| (4/3)x^2 + (4/3)x + 1/3 + 1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C = = (1/2)*ln| (4/3)x^2 + (4/3)x + 4/3| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C = = (1/2)*ln|(4/3)(x^2 + x + 1)| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C = = (1/2)*[ ln|4/3| + ln|x^2 + x + 1| ] + sqrt(3)*arctan((2x+1)/sqrt(3)) + C = = (1/2)*ln|4/3| + (1/2)*ln|x^2 + x + 1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C = = (1/2)*ln|x^2 + x + 1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C + (1/2)*ln|4/3| = = (1/2)*ln|x^2 + x + 1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C' (donde C'=C+(1/2)*ln|4/3| también es una constante y tiene derivada cero) Te dejo a ti comprobar si tu solución es correcta ;-D. Por cierto, mi solución parte de una premisa que di por supuesta, que es querer hacer la sustitución u=x+1/2 o u=(2x+1)/sqrt(3) desde el principio y luego separar las integrales según los sumandos del numerador. Sin embargo yo recomiendo separar las integrales primero, prepararlas para la sustitución y luego aplicar la sustitución. Te escribo aquí cómo habría resuelto yo esta integral desde el principio: Integral de (x+2)/(x^2+x+1) dx = = (1/2)*Integral de (2x+4)/(x^2+x+1) dx = = (1/2)*Integral de (2x+1+3)/(x^2+x+1) dx = = (1/2)*Integral de [ (2x+1)/(x^2+x+1) + 3/(x^2+x+1) ] dx = = (1/2)*Integral de (2x+1)/(x^2+x+1) dx + (3/2)*Integral de 1/(x^2+x+1) dx = (x+1/2)^2 = x^2 + x + 1/4 (x+1/2)^2 + 3/4 = x^2 + x + 1/4 + 3/4 (x+1/2)^2 + 3/4 = x^2 + x + 1 (3/4)[ (4/3)((x+1/2)^2) + 1 ] = x^2 + x + 1 (3/4)[ ( (2/sqrt(3))*(x+1/2) )^2 + 1 ] = x^2 + x + 1 (3/4)[ ((2x+1)/sqrt(3))^2 + 1 ] = x^2 + x + 1 = (1/2)*Integral de (2x+1)/(x^2+x+1) dx + (3/2)*Integral de 1/(3/4)[ ((2x+1)/sqrt(3))^2 + 1 ] dx = = (1/2)*Integral de 1/(x^2+x+1) (2x+1)dx + (3/2)*(4/3)Integral de 1/[ ((2x+1)/sqrt(3))^2 + 1 ] dx = = (1/2)*Integral de 1/(x^2+x+1) (2x+1)dx + 2*Integral de 1/[ ((2x+1)/sqrt(3))^2 + 1 ] dx = Sustitución: u = (2x+1)/sqrt(3) du = (2/sqrt(3))dx ==> (sqrt(3)/2)du = dx v = x^2+x+1 dv = (2x+1)dx = (1/2)*Integral de 1/v dv + 2*Integral de 1/(u^2 + 1) (sqrt(3)/2)du = = (1/2)*Integral de 1/v dv + 2*(sqrt(3)/2)*Integral de 1/(u^2 + 1) du = = (1/2)*Integral de 1/v dv + sqrt(3)*Integral de 1/(u^2 + 1) du = = (1/2)*ln|v| + sqrt(3)*arctan(u) = = (1/2)*ln|x^2+x+1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C Un saludo, espero que estés aprendiendo mucho en este canal! 💪
@@IntegralsForYou Yh thank you your channel really helped me thorough my quiz I think I only failed 1 out 15 integral questions and sadly it was x^3+1 I knew I had to use partial fraction just didn’t know the expansion 😭😭😭😭Thank you
Hi PARAMETATRONIC! When we have P(x) = x^n - 1 (n=3 in our case) we will always find the factor (x-1) because P(1)= 1^n - 1 = 1-1 = 0. x^2-1=(x-1)(x+1) x^3-1=(x-1)(x^2+x+1) x^4-1=(x-1)(x+1)(x^2+1) ...
Hola Julio Cesar, en este paso extraigo factor común de 3/4 para poder tener 1+(algo)^2, ya que la integral de 1+(algo)^2 es directa si el "algo" es sencillo (polinomio de primer grado). No sé si me explico bien y de hecho no sé qué es lo que no entiendes exactamente de este paso. Si aun tienes dudas, me lo podrías preguntar con algún detalle más? Gracias!
Hi! We have to calculate the integral of (x+2)/(x^2+x+1). In order to do so, we need the derivative of the denominator x^2+x+1 on the numerator. Since the numerator is x+2 and we want 2x+1, we multiply by 2 inside and divide by 2 outside the integral: (1/3)*Integral of (x+2)/(x^2+x+1) dx = = (1/3)*(1/2)*Integral of 2(x+2)/(x^2+x+1) dx = = (1/6)*Integral of (2x+4)/(x^2+x+1) dx = = (1/6)*Integral of (2x+1+3)/(x^2+x+1) dx = = (1/6)*Integral of [(2x+1)/(x^2+x+1) + 3/(x^2+x+1)] dx = = (1/6)*[ Integral of (2x+1)/(x^2+x+1) dx + 3*Integral of 1/(x^2+x+1) dx ] = = (1/6)*Integral of (2x+1)/(x^2+x+1) dx + (3/6)*Integral of 1/(x^2+x+1) dx = ... Hope it helped! ❤
Hi Derek! Here it is: Integral of 4/(x^3-4) dx = = 4*Integral of 1/(x^3-4) dx = = 4*Integral of 1/4[(x^3)/4 - 1] dx = = (4/4)*Integral of 1/[( x/(4^(1/3)) )^3 - 1] dx = = Integral of 1/[ ( x/(4^(1/3)) )^3 - 1] dx = Substitution: u = x/(4^(1/3)) = [1/(4^(1/3))]x du = 1/(4^(1/3)) dx ==> (4^(1/3)) du = dx = Integral of 1/(u^3 - 1) (4^(1/3)) du = = (4^(1/3))*Integral of 1/(u^3 - 1) du = = (4^(1/3)) * ua-cam.com/video/gGru457Jc8Y/v-deo.html = = (4^(1/3)) * [ (1/3)ln|u-1| - (1/6)ln|u^2 + u + 1| - (sqrt(3)/3)arctan( (2/sqrt(3))u + 1/sqrt(3) ) ] = = (4^(1/3)) * [ (1/3)ln|x/(4^(1/3))-1| - (1/6)ln|(x/(4^(1/3)))^2 + x/(4^(1/3)) + 1| - (sqrt(3)/3)arctan( (2/sqrt(3))x/(4^(1/3)) + 1/sqrt(3) ) ] + C The integral posted by Jashandeep we only have to multiply by -1 the integral of 1/(x^3-1): Integral of 1/(1-x^3) dx = = - Integral of 1/-(1-x^3) dx = = - Integral of 1/(-1+x^3) dx = = - Integral of 1/(x^3 -1) dx = = - ua-cam.com/video/gGru457Jc8Y/v-deo.html = = - [ (1/3)ln|x-1| - (1/6)ln|x^2+x+1| - (sqrt(3)/3)arctan((2/sqrt(3)x + 1/sqrt(3))) ] = = - (1/3)ln|x-1| + (1/6)ln|x^2+x+1| + (sqrt(3)/3)arctan((2/sqrt(3)x + 1/sqrt(3))) + C ;-D
Well, in this case you don't need Ruffini's rule if you know a^3-b^3 = (a-b)(a^2+ab+b^2). However, in general, you don't need to know a^3-b^3=(a-b)(a^2+ab+b^2) if you know the Ruffini's rule. And you don't have to know all formulas to kind of polynomial. If it was another polynomial your formula doesn't work but Ruffini's rule still do...!
I'm pretty sure you do need substitution in the last step, since the integral wasn't quite in the form you mentioned. Why am I responding to a 3 yr old comment lol?
Ah, I didn't think of turning 1/x^2+x+1 into completed square form. That's makes things much nicer. Maybe I should've kept at it... Anyways, thanks for the insight! Also, doesn't this mean you can do the same for the reciprocal of any factorable cubic? If so, that's a pretty cool family of algebraically intense integrals.
Hi! You have to do partial fraction decomposition: 1/(x^3 - 1)^2 = = 1/((x-1)(x^2+x+1))^2 = = 1/(x-1)^2(x^2+x+1)^2 = = A/(x-1) + B/(x-1)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2 = = ... As you can see, it is very long to write it in a comment. Now you have to find A,B,C,D,E,F and integrate each member of A/(x-1) + B/(x-1)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2
Hi Jashandeep singh! You only have to multiply by -1: Integral of 1/(1-x^3) dx = = - Integral of 1/-(1-x^3) dx = = - Integral of 1/(-1+x^3) dx = = - Integral of 1/(x^3 -1) dx = = - ua-cam.com/video/gGru457Jc8Y/v-deo.html = = - [ (1/3)ln|x-1| - (1/6)ln|x^2+x+1| - (sqrt(3)/3)arctan((2/sqrt(3)x + 1/sqrt(3))) ] = = - (1/3)ln|x-1| + (1/6)ln|x^2+x+1| + (sqrt(3)/3)arctan((2/sqrt(3)x + 1/sqrt(3))) + C
Hi! Because we need to express x^2 in terms of u: Integral of 1/(x^3+1) dx = Substitution: u = x^3+1 ==> u-1 = x^3 ==> (u-1)^(1/3) = x ==> (u-1)^(2/3) = x^2 du = (3x^2)dx = (3(u-1)^(2/3))dx ==> du/(u-1)^(2/3) = dx = Integral of 1/u du/(u-1)^(2/3) = = Integral of 1/[u*(u-1)^(2/3)] du = = ... Hope it helped!! 💪
@@IntegralsForYou Well there is a solution for this integral namely the following: If you don't know about Ruffini's rule or a^3-b^3 then you can do this: x^3-1=0---> x^3=1 then x=1, this is one real root, so you divide x^3-1 by x-1 in a long division to get the factor which only has complex roots: x-1/x^3-1 I x^2+x+1 x^3-x^2 ------------(-) x^2-1 x^2-x -----------(-) x-1 x-1 -----(-) 0. Now we have found the factor for x^3-1=(x-1)(x^2+x+1). I know that you don't need this for this simple integral but long division has helped me out for more complicated ones.
Sustitución: u = x+1 du = dx Integral de ln^2(x+1) dx = = Integral de ln^2(u) du = = ua-cam.com/video/NZSagniItOs/v-deo.html = = u*ln^2(u) - 2u*ln(u) + 2u = = (x+1)*ln^2(x+1) - 2(x+1)ln(x+1) + 2(x+1) + C Se puede hacer sin hacer la primera sustitución, pero es más agradable trabajar con u que con x+1. Un saludo!
Ahora no tengo mucho tiempo para una respuesta completa. Pienso en dos opciones: (1) x^2-x = (x-1/2)^2 - 1/4 y haciendo sustitución trigonométrica x-1/2 = a/cos(u) determinando "a" según 1/4 (si fuera 1/sqrt(x^2-1) sería x=1/cos(u)). (2) sqrt(x^2-x) = sqrt(x(x-1)) = sqrt(x)sqrt(x-1) . Se puede intentar el cambio u=sqrt(x) o u=sqrt(x-1). Mañana te digo algo más concreto, pero si me lo dices antes tú pues mejor :-D Un saludo Bastián!
Integral de 1/sqrt(x^2-x) dx = = Integral de 1/sqrt( (x-1/2)^2 - 1/4 ) dx = = Integral de 1/sqrt( (1/4)(2x-1)^2 - 1 ) dx = = Integral de 1/(1/2)sqrt((2x-1)^2 - 1 ) dx = = 2*Integral de 1/sqrt((2x-1)^2 - 1 ) dx = Sustitución: 2x-1 = 1/cos(u) ===> cos(u) = 1/(2x-1) ==> u = arccos(1/(2x-1)) 2 dx = sin(u)/cos^2(u) du ===> dx = (1/2) sin(u)/cos^2(u) du = 2*Integral de 1/sqrt(1/cos^2(u) - 1 ) (1/2) sin(u)/cos^2(u) du = = Integral de 1/sqrt(1/cos^2(u) - cos^2(u)/cos^2(u) ) sin(u)/cos^2(u) du = = Integral de 1/sqrt( (1-cos^2(u))/cos^2(u) ) sin(u)/cos^2(u) du = = Integral de 1/sqrt( sin^2(u)/cos^2(u) ) sin(u)/cos^2(u) du = = Integral de 1/( sin(u)/cos(u) ) sin(u)/cos^2(u) du = = Integral de cos(u)/sin(u) sin(u)/cos^2(u) du = = Integral de 1/cos(u) du = = ua-cam.com/video/qMUmvfIoTsI/v-deo.html = = ln| 1/cos(u) + tan(u) | = = ln| 1/cos(arccos(1/(2x-1))) + tan(arccos(1/(2x-1))) | = = ln| 2x - 1 + tan(arccos(1/(2x-1))) | +C
Integrals ForYou si hubiera hecho en vez de lo que hiciste usar (2x-1)=sec(u). ¿Cómo quedaría lo demás?. Sé que al final es lo mismo pero para mí hubiera sido más fácil usando que sec^2(u)-1=(2x-1)^2-1 Y luego que Tan^2(u)=(2x-1)^2-1 ¿Podrías por favor resolverlo de esa manera? (Si es qué Es posible) Esque como te digo eso se me hubiera ocurrido a mi y quiero saber si es válido igual para este ejercicio. ¿Por favor lo puedes resolver usando tan^2(u)=(2x-1)-1?
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Good work soldier. We commend your bravery.
;-D
Once you have this anti derivative, then through a very easy u-sub you can then obtain the anti derivative for 1/(x^3-a^3) in terms of "a". Saves a lot of work.
Agree! 😎
Lástima que no me topé con este video antes de mi práctica calificada del sábado pasado.
Una lástima sí... 🥲
thanks men , you realy help me :)
You're welcome! ❤
Thanks...i had forgotten this...recap
:D
I tried reading through the comments to see if you already answered these questions, but I didn't see it asked.
-@05:47 following the (*), how did you get a (1/6) coefficient and (2x+4) numerator?
-How did you know to split the 4 to (1+3)?
-How was I suppose to know to work with (x+(1/2))^2? Was it experience in re-writing denominators like (x^2+x+1)?
Also just want to thank you creating this UA-cam channel!
It's very direct, and it's been helping me through my Cal II class.
Hi! First of all, thank you for the last part of your comment, it really motivates me to keep this channel growing.
Ok, let's see:
From 5:47 we have (1/3)*Integral of (x+2)/(x^2+x+1) and we want to calculate it.
The first thing we want is to have the derivative of x^2+x+1 on the numerator and if we are lucky, just by multiplying inside and outside the integral we can get the derivative on the numerator.
f(x)=x^2+x+1 ==> f'(x)=2x+1 . Since we have "x" and not "2x", we multiply/divide inside/outside the integral by 2:
(1/3)*Integral of (x+2)/(x^2+x+1) dx =
= (1/3)*(1/2)*Integral of (2x+4)/(x^2+x+1) dx = ...
Now we have 2x+1 instead of 2x+4. Then, we can do 2x+4=2x+1+3 and split the 3 into another fraction:
= (1/6)*Integral of (2x+4)/(x^2+x+1) dx =
= (1/6)*Integral of (2x+1+3)/(x^2+x+1) dx =
= (1/6)*[Integral of (2x+1)/(x^2+x+1) dx + 3*Integral of 1/(x^2+x+1) dx ] = ...
Now we have 1/(x^2+x+1) which obviously does not have 2x+1 on the numerator and it will be impossible to get it. Then we need to write x^2+x+1 as 1+(something)^2 in order to use the arctan formula:
x^2+x+1
a^2+2ab+b^2
==>
a^2 = x^2 ==> a = x
2ab = x ==> 2xb = x ==> 2b = 1 ==> b = 1/2
Let's see what happens if we do (x+1/2)^2:
(x+1/2)^2 = x^2 + x + 1/4 =====then====> 3/4 + (x+1/2)^2 = x^2 + x + 1
I did the video of 1/(x^2+x+1) here if you want to take a look ;) ua-cam.com/video/hoaCmqkp094/v-deo.html
Hope it helped! ;-)
kannst du bitte mir zeigen wie du x^3-1 mit deine Tabelle aufgespaltet hast ?
(0:10 minute)
Hi! In the video I use the Ruffini's rule which is used when we want to divide a polynomial by x-a (where "a" is a constant): en.wikipedia.org/wiki/Ruffini's_rule
Thank you so much you know what I face great difficulty in doing the sums of integration.Can you suggest a way to be better at it?
Hi! I would say practice a lot asking and answering by yourself why we are doing each step. At the end you will see that we don't use a lot of tricks to solve integrals...
At 10:00 how did you multiply in 4/3 without foiling the whole expression out? Is there some sort of short cut to use in cases like this?
Hi Lindsey Wigand! I don't know if I understand your question but I'll explain this step with more detail:
We need something like 1/(1+(sth)^2) and we have this 4/3 multiplying, so we need to put it inside the square. So:
(1/2)Integral of 1/[ (3/4)(1+(4/3)(x + 1/2)^2 ) ] dx =
= (1/2)Integral of 4/[ 3(1+(4/3)(x + 1/2)^2 ) ] dx =
= (1/2)(4/3)Integral of 1/[ 1+(4/3)(x + 1/2)^2 ] dx =
= (2/3)Integral of 1/[ 1+ ((2/sqrt(3))^2)(x + 1/2)^2 ] dx =
= (2/3)Integral of 1/[ 1+ ( (2/sqrt(3))(x + 1/2) )^2 ] dx =
= (2/3)Integral of 1/[ 1+ ( (2/sqrt(3))x + (2/sqrt(3))(1/2) )^2 ] dx =
= (2/3)Integral of 1/[ 1+ ( (2/sqrt(3))x + 1/sqrt(3) )^2 ] dx = ...
I did something like:
(a/b)(c+d)^2 =
= ((sqrt(a)/sqrt(b))^2)(c+d)^2 =
= [ (sqrt(a)/sqrt(b))(c+d) ]^2 =
= [ (sqrt(a)/sqrt(b))c + (sqrt(a)/sqrt(b))d ]^2
Hope it helped... and if not, could you please ask me again with more details about your question? Thanks! ;-D
Makes much more sense now, thank you!
You're welcome!! ;-D
May you please try to explain verbally too,coz I don't understand so many things
Hi, I know that it would be easier for you if I talked but I prefer to let you think why am I doing each step and if you don't understand it you can ask me in a comment. In my opinion, it is the best way to practice integration 😊
Are these the arcane runes?
Hola, cuando tienes la integral de (x+2) / ((x+1/2)^2 + 3/4) , ¿es posible sustituir u = x+ 1/2 para después integrar por sustitución trigonométrica?
Por supuesto! Aunque quizás yo aprovecharía e intentaría incluir el 3/4 en la sustitución con "u":
Integral de (x+2)/[ (x+1/2)^2 + 3/4 ] dx =
= Integral de (x+2)/(3/4)*[ (4/3)(x+1/2)^2 + 1 ] dx =
= (4/3)*Integral de (x+2)/[ ( (2/sqrt(3))(x+1/2))^2 + 1 ] dx =
= (4/3)*Integral de (x+2)/[ ( (2/sqrt(3))x + 1/sqrt(3) )^2 + 1 ] dx =
Sustitución:
u = (2/sqrt(3))x + 1/sqrt(3) = (2x+1)/sqrt(3) ==> (sqrt(3)u-1)/2 = x
du = (2/sqrt(3))dx ==> (sqrt(3)/2)du = dx
= (4/3)*Integral de ((sqrt(3)u-1)/2 + 2)/(u^2 + 1) (sqrt(3)/2)du =
= (4/3)*(sqrt(3)/2)*Integral de ( (sqrt(3)/2)u - 1/2 + 2)/(u^2 + 1) du =
= (2/sqrt(3))*Integral de ( (sqrt(3)/2)u + 3/2)/(u^2 + 1) du = ...
Espero haberte ayudado! Un saludo!
@@IntegralsForYou Sí, muchas gracias por darte el tiempo de responder. Si no es mucha molestia, el resultado de la integral, es - ln| 1 / sqrt(u^2 +1 ) | + sqrt(3) arctan(u) y realizando la sustitución de u, entonces es -ln| 1 / [sqrt( (2/sqrt(3) x + 1/sqrt(3) )^2 +1 )] | + arctan(2/sqrt(3) x + 1/sqrt(3) )^2 , cierto?
@@joseorduna7010
Hola! Hoy tengo algo más de tiempo para poder completar mi respuesta anterior:
Integral de (x+2)/[ (x+1/2)^2 + 3/4 ] dx =
= Integral de (x+2)/(3/4)*[ (4/3)(x+1/2)^2 + 1 ] dx =
= (4/3)*Integral de (x+2)/[ ( (2/sqrt(3))(x+1/2))^2 + 1 ] dx =
= (4/3)*Integral de (x+2)/[ ( (2/sqrt(3))x + 1/sqrt(3) )^2 + 1 ] dx =
Sustitución:
u = (2/sqrt(3))x + 1/sqrt(3) = (2x+1)/sqrt(3) ==> (sqrt(3)u-1)/2 = x
du = (2/sqrt(3))dx ==> (sqrt(3)/2)du = dx
= (4/3)*Integral de ((sqrt(3)u-1)/2 + 2)/(u^2 + 1) (sqrt(3)/2)du =
= (4/3)*(sqrt(3)/2)*Integral de ( (sqrt(3)/2)u - 1/2 + 2)/(u^2 + 1) du =
= (2/sqrt(3))*Integral de ( (sqrt(3)/2)u + 3/2)/(u^2 + 1) du =
= (2/sqrt(3))*[ (sqrt(3)/2)*Integral de u/(u^2+1) du + (3/2)*Integral de 1/(u^2 + 1) du ] =
= Integral de u/(u^2+1) du + sqrt(3)*Integral de 1/(u^2+1) du =
= (1/2)*Integral de 2u/(u^2+1) du + sqrt(3)*Integral de 1/(u^2+1) du =
= (1/2)*ln|u^2+1| + sqrt(3)*arctan(u) =
= (1/2)*ln| [(2/sqrt(3))x + 1/sqrt(3)]^2 + 1| + sqrt(3)*arctan((2/sqrt(3))x + 1/sqrt(3)) + C =
Hasta aquí podríamos decir que ya hemos acabado, pero simplifiquemos un poco, no? ;-)
= (1/2)*ln| (4/3)x^2 + (4/3)x + 1/3 + 1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C =
= (1/2)*ln| (4/3)x^2 + (4/3)x + 4/3| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C =
= (1/2)*ln|(4/3)(x^2 + x + 1)| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C =
= (1/2)*[ ln|4/3| + ln|x^2 + x + 1| ] + sqrt(3)*arctan((2x+1)/sqrt(3)) + C =
= (1/2)*ln|4/3| + (1/2)*ln|x^2 + x + 1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C =
= (1/2)*ln|x^2 + x + 1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C + (1/2)*ln|4/3| =
= (1/2)*ln|x^2 + x + 1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C' (donde C'=C+(1/2)*ln|4/3| también es una constante y tiene derivada cero)
Te dejo a ti comprobar si tu solución es correcta ;-D. Por cierto, mi solución parte de una premisa que di por supuesta, que es querer hacer la sustitución u=x+1/2 o u=(2x+1)/sqrt(3) desde el principio y luego separar las integrales según los sumandos del numerador. Sin embargo yo recomiendo separar las integrales primero, prepararlas para la sustitución y luego aplicar la sustitución. Te escribo aquí cómo habría resuelto yo esta integral desde el principio:
Integral de (x+2)/(x^2+x+1) dx =
= (1/2)*Integral de (2x+4)/(x^2+x+1) dx =
= (1/2)*Integral de (2x+1+3)/(x^2+x+1) dx =
= (1/2)*Integral de [ (2x+1)/(x^2+x+1) + 3/(x^2+x+1) ] dx =
= (1/2)*Integral de (2x+1)/(x^2+x+1) dx + (3/2)*Integral de 1/(x^2+x+1) dx =
(x+1/2)^2 = x^2 + x + 1/4
(x+1/2)^2 + 3/4 = x^2 + x + 1/4 + 3/4
(x+1/2)^2 + 3/4 = x^2 + x + 1
(3/4)[ (4/3)((x+1/2)^2) + 1 ] = x^2 + x + 1
(3/4)[ ( (2/sqrt(3))*(x+1/2) )^2 + 1 ] = x^2 + x + 1
(3/4)[ ((2x+1)/sqrt(3))^2 + 1 ] = x^2 + x + 1
= (1/2)*Integral de (2x+1)/(x^2+x+1) dx + (3/2)*Integral de 1/(3/4)[ ((2x+1)/sqrt(3))^2 + 1 ] dx =
= (1/2)*Integral de 1/(x^2+x+1) (2x+1)dx + (3/2)*(4/3)Integral de 1/[ ((2x+1)/sqrt(3))^2 + 1 ] dx =
= (1/2)*Integral de 1/(x^2+x+1) (2x+1)dx + 2*Integral de 1/[ ((2x+1)/sqrt(3))^2 + 1 ] dx =
Sustitución:
u = (2x+1)/sqrt(3)
du = (2/sqrt(3))dx ==> (sqrt(3)/2)du = dx
v = x^2+x+1
dv = (2x+1)dx
= (1/2)*Integral de 1/v dv + 2*Integral de 1/(u^2 + 1) (sqrt(3)/2)du =
= (1/2)*Integral de 1/v dv + 2*(sqrt(3)/2)*Integral de 1/(u^2 + 1) du =
= (1/2)*Integral de 1/v dv + sqrt(3)*Integral de 1/(u^2 + 1) du =
= (1/2)*ln|v| + sqrt(3)*arctan(u) =
= (1/2)*ln|x^2+x+1| + sqrt(3)*arctan((2x+1)/sqrt(3)) + C
Un saludo, espero que estés aprendiendo mucho en este canal! 💪
sos un dios perro
!!!
jajaja gracias...!
Thanks a lot for Uploading this video
My pleasure! 💪😎
please what was the beginning thing you did?
Hi! We factorized x^3-1 into (x-1)(x^2+x+1) using the Ruffini's rule: en.wikipedia.org/wiki/Ruffini%27s_rule 😉
@@IntegralsForYou ok thank you 🙏
@@kidd1941 My pleasure! ❤
@@IntegralsForYou Yh thank you your channel really helped me thorough my quiz I think I only failed 1 out 15 integral questions and sadly it was x^3+1 I knew I had to use partial fraction just didn’t know the expansion 😭😭😭😭Thank you
@@kidd1941 Next time you will get 15/15 💪❤
where are you from?
Hi! I'm from Spain 😊
Thank you for the demo! Btw how do you know that you have to extract (x-1) from (x^3 - 1) in order to factorise?
Hi PARAMETATRONIC! When we have P(x) = x^n - 1 (n=3 in our case) we will always find the factor (x-1) because P(1)= 1^n - 1 = 1-1 = 0.
x^2-1=(x-1)(x+1)
x^3-1=(x-1)(x^2+x+1)
x^4-1=(x-1)(x+1)(x^2+1)
...
Brilliant! Cheers!
:-D
muchas gracias me ayudo mucho..!!! pero tengo una duda en el minuto 9:10 , por que aumentas 1 + 4/3?
Hola Julio Cesar, en este paso extraigo factor común de 3/4 para poder tener 1+(algo)^2, ya que la integral de 1+(algo)^2 es directa si el "algo" es sencillo (polinomio de primer grado). No sé si me explico bien y de hecho no sé qué es lo que no entiendes exactamente de este paso. Si aun tienes dudas, me lo podrías preguntar con algún detalle más? Gracias!
Thank you so much.
:-D
how about 1/(x^3 - 1)^2
Hi! In this case:
1/(x^3-1)^2 = 1/[(x-1)(x^2+x+1)]^2 = 1/[(x-1)^2 (x^2+x+1)^2] = A/(x-1) + B/(x-1)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2 = ...
cannot for the life of me figure out where you got the 1/6 coefficient. Anyone know?
Hi! We have to calculate the integral of (x+2)/(x^2+x+1). In order to do so, we need the derivative of the denominator x^2+x+1 on the numerator. Since the numerator is x+2 and we want 2x+1, we multiply by 2 inside and divide by 2 outside the integral:
(1/3)*Integral of (x+2)/(x^2+x+1) dx =
= (1/3)*(1/2)*Integral of 2(x+2)/(x^2+x+1) dx =
= (1/6)*Integral of (2x+4)/(x^2+x+1) dx =
= (1/6)*Integral of (2x+1+3)/(x^2+x+1) dx =
= (1/6)*Integral of [(2x+1)/(x^2+x+1) + 3/(x^2+x+1)] dx =
= (1/6)*[ Integral of (2x+1)/(x^2+x+1) dx + 3*Integral of 1/(x^2+x+1) dx ] =
= (1/6)*Integral of (2x+1)/(x^2+x+1) dx + (3/6)*Integral of 1/(x^2+x+1) dx = ...
Hope it helped! ❤
How would you solve the integral 4/x^3-4? as well as the integral posted by Jashandeep? Thank you
Hi Derek! Here it is:
Integral of 4/(x^3-4) dx =
= 4*Integral of 1/(x^3-4) dx =
= 4*Integral of 1/4[(x^3)/4 - 1] dx =
= (4/4)*Integral of 1/[( x/(4^(1/3)) )^3 - 1] dx =
= Integral of 1/[ ( x/(4^(1/3)) )^3 - 1] dx =
Substitution:
u = x/(4^(1/3)) = [1/(4^(1/3))]x
du = 1/(4^(1/3)) dx ==> (4^(1/3)) du = dx
= Integral of 1/(u^3 - 1) (4^(1/3)) du =
= (4^(1/3))*Integral of 1/(u^3 - 1) du =
= (4^(1/3)) * ua-cam.com/video/gGru457Jc8Y/v-deo.html =
= (4^(1/3)) * [ (1/3)ln|u-1| - (1/6)ln|u^2 + u + 1| - (sqrt(3)/3)arctan( (2/sqrt(3))u + 1/sqrt(3) ) ] =
= (4^(1/3)) * [ (1/3)ln|x/(4^(1/3))-1| - (1/6)ln|(x/(4^(1/3)))^2 + x/(4^(1/3)) + 1| - (sqrt(3)/3)arctan( (2/sqrt(3))x/(4^(1/3)) + 1/sqrt(3) ) ] + C
The integral posted by Jashandeep we only have to multiply by -1 the integral of 1/(x^3-1):
Integral of 1/(1-x^3) dx =
= - Integral of 1/-(1-x^3) dx =
= - Integral of 1/(-1+x^3) dx =
= - Integral of 1/(x^3 -1) dx =
= - ua-cam.com/video/gGru457Jc8Y/v-deo.html =
= - [ (1/3)ln|x-1| - (1/6)ln|x^2+x+1| - (sqrt(3)/3)arctan((2/sqrt(3)x + 1/sqrt(3))) ] =
= - (1/3)ln|x-1| + (1/6)ln|x^2+x+1| + (sqrt(3)/3)arctan((2/sqrt(3)x + 1/sqrt(3))) + C
;-D
Awesome, thank you much! :-)
You're welcome!! ;-D
No need for Ruffini's rule. We know a^3-b^3 = (a-b)(a^2+ab+b^2)
No need of substitution in the last step. We know that
∫ dx/(a^2+X^2) = 1/a Tan^-1 x/a
Well, in this case you don't need Ruffini's rule if you know a^3-b^3 = (a-b)(a^2+ab+b^2).
However, in general, you don't need to know a^3-b^3=(a-b)(a^2+ab+b^2) if you know the Ruffini's rule. And you don't have to know all formulas to kind of polynomial. If it was another polynomial your formula doesn't work but Ruffini's rule still do...!
I'm pretty sure you do need substitution in the last step, since the integral wasn't quite in the form you mentioned. Why am I responding to a 3 yr old comment lol?
@@IntegralsForYou please explain ruffinis rule
I want x cube + 1/x dx plz can u keep it
Integral of (x^3+1)/x dx =
= Integral of ( x^3/x + 1/x ) dx =
= Integral of ( x^2 + 1/x ) dx =
= x^3/3 + ln|x| + C
😉
you are crazy :)
hahaha calm down....
Ah, I didn't think of turning 1/x^2+x+1 into completed square form. That's makes things much nicer. Maybe I should've kept at it... Anyways, thanks for the insight! Also, doesn't this mean you can do the same for the reciprocal of any factorable cubic? If so, that's a pretty cool family of algebraically intense integrals.
1dx/(x³-1)^². Please help
Hi! You have to do partial fraction decomposition:
1/(x^3 - 1)^2 =
= 1/((x-1)(x^2+x+1))^2 =
= 1/(x-1)^2(x^2+x+1)^2 =
= A/(x-1) + B/(x-1)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2 =
= ...
As you can see, it is very long to write it in a comment. Now you have to find A,B,C,D,E,F and integrate each member of A/(x-1) + B/(x-1)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2
Is there any other way?
This is way too lengthy
@@amber1171 Maybe there is another way but I only know this one...
So nice ❤️❤️❤️❤️❤️
One of my favorites! ;-D
isnt isnt 2sqrt(3)/3 arctan (..) not sqrt(3)/3
Hi! In the last step we have (2/3)*(sqrt(3)/2) which is sqrt(3)/3... I'm not sure if you were asking for this one...
Hi! In the last step we have (2/3)*(sqrt(3)/2) which is sqrt(3)/3... I'm not sure if you were asking for this one...
Mrc
shouldn't that be Ax^2+Ax+A+Bx-Cx-C
Thanks but please solve the integral of 1/1-x^3
Hi Jashandeep singh! You only have to multiply by -1:
Integral of 1/(1-x^3) dx =
= - Integral of 1/-(1-x^3) dx =
= - Integral of 1/(-1+x^3) dx =
= - Integral of 1/(x^3 -1) dx =
= - ua-cam.com/video/gGru457Jc8Y/v-deo.html =
= - [ (1/3)ln|x-1| - (1/6)ln|x^2+x+1| - (sqrt(3)/3)arctan((2/sqrt(3)x + 1/sqrt(3))) ] =
= - (1/3)ln|x-1| + (1/6)ln|x^2+x+1| + (sqrt(3)/3)arctan((2/sqrt(3)x + 1/sqrt(3))) + C
a^3- b^3 = (a - b)(a^2+ ab + b^2 )
⇒ 1 - x^3= (1 - x)(1 + x + x^2 )= (1 - x)(x^2+ x + 1)
1/((1 - x^3 ) )=1/(1 - x)(x^2+ x + 1)
Using partial fraction let
1/((1 - x^3 ) )=1/(1 - x)(x^2+ x + 1) =A/(1-x)+(B x +C)/(x^2+ x + 1)
1 = A (x^2+ x + 1)+B x (1 - x)+ C (1 - x) by putting x = 1, A =1/3
Comparing coefficients of x on both sides
A - B = 0 Hence B = A =1/3
A + C = 1 Hence 1/3+ C = 1 ⇒ C = 1 -1/3 = 2/3
B x + C =1/3 x +2/3=(x + 2)/3
Therefore 1/(1 - x^3 )=1/3(1 - x) +(x + 2)/3(x^2+ x + 1)
=1/3(1 - x) +(2x + 4)/6(x^2+ x + 1)
=1/3(1 - x) +(2x + 1 + 3)/6(x^2+ x + 1)
=1/3(1 - x) +(2x + 1)/6(x^2+ x + 1) +1/2(x^2+ x + 1)
=1/3(1 - x) +(2x + 1)/6(x^2+ x + 1) + ½{1/((x + ½)^2 + (√3/2)^2 )}
Hence ∫1/((1 - x^3 ) ) d x =1/3 ∫1/(1-x) d x +1/6 ∫(2 x + 1)/((x^2+ x + 1) ) d x +1/2 ∫1/((x + ½)^2 + (√3/2)^2 ) d x
= - (1 )/3 ln(1 - x) + (1 )/6 ln(x^2 + x + 1)+ (1/2) / (√3 /2) {tan^(-1)(x + ½)/√3 /2} + C
= - 1/3 ln(1 - x) + 1/6 ln(x^2+ x + 1) + (1/√3 ) {tan^(-1)(2x + 1)/√3 } + C
= 1/6 {ln(x^2 + x + 1) - 2 ln(1 - x) + 2√3 tan^(-1)(2x + 1)/√3} + C
why isnt it possible? integral of 1/x^3+1 = 1/3^2 * ln|x^3 + 1| + C. Considering dx = du/3x^2 and u = x^3+1.
Hi! Because we need to express x^2 in terms of u:
Integral of 1/(x^3+1) dx =
Substitution:
u = x^3+1 ==> u-1 = x^3 ==> (u-1)^(1/3) = x ==> (u-1)^(2/3) = x^2
du = (3x^2)dx = (3(u-1)^(2/3))dx ==> du/(u-1)^(2/3) = dx
= Integral of 1/u du/(u-1)^(2/3) =
= Integral of 1/[u*(u-1)^(2/3)] du =
= ...
Hope it helped!! 💪
Thanks
You're welcome! ;-D
Can you solve this integral 2cot²+cos²3x
Hi Xzel 10! Here you have the answer:
cos^2(x) = (1/2)(1+cos(2x))
cos^2(3x)=(1/2)(1+cos(6x))
cot^2(x) = cos^2(x)/sin^2(x) = (1-sin^2(x))/sin^2(x) = 1/sin^2(x) - sin^2(x)/sin^2(x) = 1/sin^2(x) - 1
Integral of [ 2cot^2(x) + cos^2(3x) ] dx =
= Integral of [ 2(1/sin^2(x) - 1) + (1/2)(1+cos(6x)) ] dx =
= Integral of [ 2/sin^2(x) - 2 + 1/2 + (1/2)cos(6x) ] dx =
= Integral of [ 2/sin^2(x) - 3/2 + (1/12)6cos(6x) ] dx =
= 2Integral of 1/sin^2(x) dx - (3/2)Integral of dx + (1/12)Integral of 6cos(6x) dx =
= 2cot(x) - (3/2)x + (1/12)sin(6x) + C
;-D
Is their any alternate method to solve this??
Hi, Vikash Yadav! I don't think so... Maybe you can do the same method differently but in the end you will be doing the same
@@IntegralsForYou Well there is a solution for this integral namely the following:
If you don't know about Ruffini's rule or a^3-b^3 then you can do this:
x^3-1=0---> x^3=1 then x=1, this is one real root, so you divide x^3-1 by x-1 in a long division to get the factor which only has complex roots:
x-1/x^3-1 I x^2+x+1
x^3-x^2
------------(-)
x^2-1
x^2-x
-----------(-)
x-1
x-1
-----(-)
0.
Now we have found the factor for x^3-1=(x-1)(x^2+x+1).
I know that you don't need this for this simple integral but long division has helped me out for more complicated ones.
Any alternate method
I am sorry, this is the only way I found to solve this problem...
¿La integral de Ln^2(x+1)? Intento hacerlo por partes pero no me saleeeee 😩
Sustitución:
u = x+1
du = dx
Integral de ln^2(x+1) dx =
= Integral de ln^2(u) du =
= ua-cam.com/video/NZSagniItOs/v-deo.html =
= u*ln^2(u) - 2u*ln(u) + 2u =
= (x+1)*ln^2(x+1) - 2(x+1)ln(x+1) + 2(x+1) + C
Se puede hacer sin hacer la primera sustitución, pero es más agradable trabajar con u que con x+1.
Un saludo!
¿Cómo saco la integral de la Arcocosecante, la arcosecante y la arcocotangente?
Por partes como arcocoseno, arcoseno y arcotangente. Tomo nota para hacerlas en mis próximos videos :-D. Un saludo Bastián!
Saludos. ¡Gracias por siempre responder!
Ojalá este semestre pase Cálculo 2 🙏
Muchos ánimos para Cálculo 2 y las demás materias que tengas!!!
¿Integral de (x^2-x)^(-0,5)?
Ahora no tengo mucho tiempo para una respuesta completa. Pienso en dos opciones:
(1) x^2-x = (x-1/2)^2 - 1/4 y haciendo sustitución trigonométrica x-1/2 = a/cos(u) determinando "a" según 1/4 (si fuera 1/sqrt(x^2-1) sería x=1/cos(u)).
(2) sqrt(x^2-x) = sqrt(x(x-1)) = sqrt(x)sqrt(x-1) . Se puede intentar el cambio u=sqrt(x) o u=sqrt(x-1).
Mañana te digo algo más concreto, pero si me lo dices antes tú pues mejor :-D
Un saludo Bastián!
Integral de 1/sqrt(x^2-x) dx =
= Integral de 1/sqrt( (x-1/2)^2 - 1/4 ) dx =
= Integral de 1/sqrt( (1/4)(2x-1)^2 - 1 ) dx =
= Integral de 1/(1/2)sqrt((2x-1)^2 - 1 ) dx =
= 2*Integral de 1/sqrt((2x-1)^2 - 1 ) dx =
Sustitución:
2x-1 = 1/cos(u) ===> cos(u) = 1/(2x-1) ==> u = arccos(1/(2x-1))
2 dx = sin(u)/cos^2(u) du ===> dx = (1/2) sin(u)/cos^2(u) du
= 2*Integral de 1/sqrt(1/cos^2(u) - 1 ) (1/2) sin(u)/cos^2(u) du =
= Integral de 1/sqrt(1/cos^2(u) - cos^2(u)/cos^2(u) ) sin(u)/cos^2(u) du =
= Integral de 1/sqrt( (1-cos^2(u))/cos^2(u) ) sin(u)/cos^2(u) du =
= Integral de 1/sqrt( sin^2(u)/cos^2(u) ) sin(u)/cos^2(u) du =
= Integral de 1/( sin(u)/cos(u) ) sin(u)/cos^2(u) du =
= Integral de cos(u)/sin(u) sin(u)/cos^2(u) du =
= Integral de 1/cos(u) du =
= ua-cam.com/video/qMUmvfIoTsI/v-deo.html =
= ln| 1/cos(u) + tan(u) | =
= ln| 1/cos(arccos(1/(2x-1))) + tan(arccos(1/(2x-1))) | =
= ln| 2x - 1 + tan(arccos(1/(2x-1))) | +C
Integrals ForYou si hubiera hecho en vez de lo que hiciste usar (2x-1)=sec(u). ¿Cómo quedaría lo demás?.
Sé que al final es lo mismo pero para mí hubiera sido más fácil usando que sec^2(u)-1=(2x-1)^2-1
Y luego que
Tan^2(u)=(2x-1)^2-1
¿Podrías por favor resolverlo de esa manera?
(Si es qué Es posible)
Esque como te digo eso se me hubiera ocurrido a mi y quiero saber si es válido igual para este ejercicio.
¿Por favor lo puedes resolver usando tan^2(u)=(2x-1)-1?
sorry I didn't foil (Bx+C)(x-1)
Hard prblm
method is good but result is not right
Hi! Did I do any mistake? In which minute?
The result was fine, just not in the exact form you might find in an online integral calculator.
Hell nah