Hi! Because the derivative of 1/x is -1/x^2 😉 Another way to solve this integral is by doing a u-substitution: Integral of e^(1/x)/x^2 dx = = Integral of e^(1/x) (1/x^2)dx = Substitution: u = 1/x du = (-1/x^2)dx ==> -du = (1/x^2)dx = Integral of e^u (-du) = = - Integral of e^u du = = -e^u = = -e^(1/x) + C Hope it helped! 💪
Hi! I just put it on my TO DO list! Meanwhile I can write here the details: Integral of e^(1/x)/x^2 dx = = Integral of e^(1/x)*(1/x^2) dx = Substitution: u = e^(1/x) du = e^(1/x)*(-1/x^2) dx ==> - du = e^(1/x)*(1/x^2) dx = Integral of (-du) = = - Integral of du = = -u = = -e^(1/x) + C ❤
jajaja uno de los mejores comentarios con diferencia! Me alegro mucho haberte ayudado! Muchas gracias por tu comentario, en serio, me alegraste el día! 🎉
Hi! Here you have the solution: Integral of e^(1/x)/x^3 dx = = Integral of (e^(1/x))*(1/x)*(1/x^2)*dx = Substitution: u = 1/x du = (-1/x^2)*dx ==> -du = (1/x^2)*dx = Integral of (e^u)*u*(-du) = = - Integral of u*e^u du = = - ua-cam.com/video/oOlGshv2ycc/v-deo.html = = - ( u*e^u - e^u ) = = (e^u)(1 - u) = = (e^(1/x))(1 - 1/x) + C Hope it helped! ;-D
Hola! La integral de e^(1/x) no puede expresarse en términos de un número finito de funciones elementales. Para este caso se debería usar algún método de aproximación de la solución de la integral
Hola Estudiante 58! De hecho u no es 1/x^2, es 1/x ya que su derivada es -1/x^2. Si quieres otra manera de ver la sustitución puede ser la sieguiente: Integral de[ e^(1/x)]/x^2 dx = = Integral de [e^(1/x)] 1/x^2 dx = Sustitución: u = 1/x du = -1/x^2 dx ==> -du = 1/x^2 dx = Integral de (e^u)(-du) = = - Integral de e^u du = = -e^u = = - e^(1/x) + C
Hi! It doesn't change a lot. Let's see: Integral of e^(2/x)/x^2 dx = = Integral of e^(2/x) (1/x^2) dx = Substitution: u = 2/x = 2x^(-1) du = -2x^(-2) dx = -2/x^2 dx ==> du/-2 = 1/x^2 dx = Integral of e^u du/-2 = = (-1/2)Integral of e^u du = = (-1/2)e^u = = (-1/2)e^(2/x) + C Hope it helped! ;-D
Hi, here you have the solution: Integral of (e^(-1/x)+1)/x^2 dx = = Integral of ( e^(-1/x)/x^2 + 1/x^2 ) dx = = Integral of e^(-1/x) (1/x^2)dx + Integral of 1/x^2 dx = Substitution: u = -1/x du = (1/x^2)dx = Integral of e^u du + Integral of x^(-2) dx = = e^u + x^(-2+1)/(-2+1) = = e^(-1/x) + x^(-1)/(-1) = = e^(-1/x) - 1/x + C
Of course! Here you have the solution: Integral of 1/(e^x-1)^2 dx = Substitution: u = e^x-1 ==> u+1 = e^x du = (e^x)dx = (u+1)dx ==> du/(u+1) = dx = Integral of 1/u^2 du/(u+1) = = Integral of 1/(u^2)(u+1) du = = ua-cam.com/video/GyL5qvNe4bg/v-deo.html = = -ln|u| - 1/u + ln|u+1| = = -ln|e^x-1| - 1/(e^x-1) + ln|e^x-1+1| = = -ln|e^x-1| - 1/(e^x-1) + ln|e^x| = = -ln|e^x-1| - 1/(e^x-1) + x + C Hope it helped! ;-D
@Mustafa Attarwala I have a TO DO list for the videos, I can add this one to the list but you will have to wait a lot of time for the video to be uploaded. I suggest you to ask here in a comment the doubts you have. If you want to send me a picture you can do it on instagram 😉
Hi! here you have the solution: Integral of x/e^(2x) dx = = Integral of x*e^(2x) dx = Parts: Integral of u dv = uv - Integral of v du u = x ==> du = dx dv = e^(-2x) dx ==> v = (-1/2)e^(-2x) = x*(-1/2)e^(-2x) - Integral of (-1/2)e^(-2x) dx = = x*(-1/2)e^(-2x) + (1/2)Integral of e^(-2x) dx = = x*(-1/2)e^(-2x) + (1/2)(-1/2)e^(-2x) = = x*(-1/2)e^(-2x) - (1/4)e^(-2x) = = (-1/4)(2x+1)e^(-2x) + C Then: Integral of x/e^(2x) dx from 0 to 1 = = (-1/4)(2*1+1)e^(-2*1) - (-1/4)(2*0+1)e^(-2*0) = = (-1/4)(3)e^(-2) - (-1/4)(1)e^0 = = (-3/4)e^(-2) + 1/4 = = 0.1484...
Hola! Como la derivada de 1/x es -1/x^2 y sólo tenemos 1/x^2 , podemos multiplicar por -1 dentro y fuera de la integral, o lo que es lo mismo, poner el signo "-" dentro y fuera de la integral. Así ya tenemos -1/x^2 que es lo que queríamos. Si lo ves más fácil te dejo otra manera de aplicar el método de sustitución para esta integral: Integral de e^(1/x)/x^2 dx = = Integral de e^(1/x) 1/x^2 dx = Sustitución: u = 1/x du = -1/x^2 dx ==> -du = 1/x^2 dx = Integral de e^u (-du) = = - Integral de e^u du = = - e^u = = - e^(1/x) + C Hay mucha gente que lo entiende mejor así. Personalmente los dos métodos me parecen igual de correctos, así que escoge! ;-d
Hi Abel! I've seen you were asking for this video and you have already found it! 😉I share the link in the other comment in case somebody wants to know it! Hope my channel is helping you! 👍
@@asterixmusic4928 Of course, here you have: Integral of (2x-4)cos(x^2 - 4x) dx = = Integral of cos(x^2 - 4x) (2x-4)dx = Substitution: u = x^2 - 4x du = (2x-4)dx = Integral of cos(u) du = = sin(u) = = sin(x^2 - 4x) + C ;-D
Hola marlyz johana burgos barrera! El -1 sale de la formula de derivación de x^n: Derivada de x^n = n*x^(n-1) Derivada de x^(-1) = (-1)x^(-1-1) = -x^(-2) = -1/x^2 Un saludo!
Hi! I don't understand which expression you are asking for: Integral of e^(1/x + 2/x^2) dx ? Integral of e^( 1/(x + 2) + 1/x^2 ) dx ? Integral of ( e^(1/x) + 2/x^2 ) dx ?
Hi suhreeen, I don't think so... If you let u=1/x^2 then you will have to do a second substitution t=sqrt(u), but it is easier to do t=1/x directly [ t=sqrt(u)=sqrt(1/x^2)=1/x ]: Integral of e^(1/x) 1/x^2 dx = Substitution: u = 1/x^2 = x^-2 ==> sqrt(u) = 1/x du = -2x^-3 dx = -2/x^3 dx ==> x/-2 du = 1/x^2 dx ==> sqrt(u)/-2 du = 1/x^2 dx = Integral of e^sqrt(u) sqrt(u)/-2 du = = (-1/2)Integral of e^sqrt(u) sqrt(u) du = Substitution: t=sqrt(u) dt = 1/2sqrt(u) du ==> 2 dt = sqrt(u) du = (-1/2)Integral of e^t 2 dt = = (-2/2)Integral of e^t dt = = - e^t = = - e^sqrt(u) = = - e^(1/x) + C Hope I answered your question. If not, ask me again with more details ;-D
Hi! Here you have the solution: Integral of e^(1/x)/x^3 dx = = Integral of (e^(1/x))*(1/x)*(1/x^2)*dx = Substitution: u = 1/x du = (-1/x^2)*dx ==> -du = (1/x^2)*dx = Integral of (e^u)*u*(-du) = = - Integral of u*e^u du = = - ua-cam.com/video/oOlGshv2ycc/v-deo.html = = - ( u*e^u - e^u ) = = (e^u)(1 - u) = = (e^(1/x))(1 - 1/x) + C Hope it helped! ;-D
![Evaluate the Integral on [1, 2] e^(1/x)/x^2 dx. Natural exponential function](/img/n.gif)
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Student, Austria
Unbeliefable how you act with
complecated maths.
Many thanks for your help
You're welcome!! Thanks for your comment! ;-D
i have a question, why did you afect all the integral whit (-)
Hi! Because the derivative of 1/x is -1/x^2 😉
Another way to solve this integral is by doing a u-substitution:
Integral of e^(1/x)/x^2 dx =
= Integral of e^(1/x) (1/x^2)dx =
Substitution:
u = 1/x
du = (-1/x^2)dx ==> -du = (1/x^2)dx
= Integral of e^u (-du) =
= - Integral of e^u du =
= -e^u =
= -e^(1/x) + C
Hope it helped! 💪
@@IntegralsForYou thanks bro, if you got a view of mine you gonna have a like too, forever
@@clr820 Thank you! If you want to help me, please share this channel with your friends! ❤
Thank you, you are god sir
Thank you! 😊
¡Excelente vídeo!
Muchas gracias, Vicente :D
Can you do same while taking u=e^1/x?
Make a video on it.
Hi! I just put it on my TO DO list! Meanwhile I can write here the details:
Integral of e^(1/x)/x^2 dx =
= Integral of e^(1/x)*(1/x^2) dx =
Substitution:
u = e^(1/x)
du = e^(1/x)*(-1/x^2) dx ==> - du = e^(1/x)*(1/x^2) dx
= Integral of (-du) =
= - Integral of du =
= -u =
= -e^(1/x) + C
❤
*_Thank bro 👍🏻. I found the solution._*
@@Yash-Class9-JEE Nice! 💪
thank you sir/ma'am :) since there's no sound :)
Since there's no sound let's say I'm both 😉😜
@@IntegralsForYou lol
Me salvaste la noche te amo, me amo bashjhsakuisdefgdgf xoxoxoxo Te debo un abrazo I love you umaaaaa
jajaja uno de los mejores comentarios con diferencia! Me alegro mucho haberte ayudado! Muchas gracias por tu comentario, en serio, me alegraste el día! 🎉
Sir how to integral e^(x^2/2)
Hi! I'm sorry but this is a non-elementary integral...
can u help me with this...how to solve x^-(1/2)*e^x^2
Hi! It is a non-elementary integral, it cannot be expressed in terms of finite standard functions...
How would I evaluate if bottom is x^3
Hi! Here you have the solution:
Integral of e^(1/x)/x^3 dx =
= Integral of (e^(1/x))*(1/x)*(1/x^2)*dx =
Substitution:
u = 1/x
du = (-1/x^2)*dx ==> -du = (1/x^2)*dx
= Integral of (e^u)*u*(-du) =
= - Integral of u*e^u du =
= - ua-cam.com/video/oOlGshv2ycc/v-deo.html =
= - ( u*e^u - e^u ) =
= (e^u)(1 - u) =
= (e^(1/x))(1 - 1/x) + C
Hope it helped! ;-D
Jajajaja pensé que era más difícil muchas gracias
jeje lo parece, eh? Pero cuando ves que la derivada de 1/x está presente se soluciona todo casi de manera mágica 😊
Thanks
My pleasure! 😉
e quanto a integral de e^(1/x)?
Hola! La integral de e^(1/x) no puede expresarse en términos de un número finito de funciones elementales. Para este caso se debería usar algún método de aproximación de la solución de la integral
@@IntegralsForYou thank you.
@kael max Anytime! 😉
Gracias!!
Un placer! 😊
Súper. Buen vídeo.
Gracias Margu! ;-D
Integration x^-4*e^(-1/x)
thanks god I find this one
Glad you found it! ❤
Porque u= 1/x^2?
Hola Estudiante 58! De hecho u no es 1/x^2, es 1/x ya que su derivada es -1/x^2. Si quieres otra manera de ver la sustitución puede ser la sieguiente:
Integral de[ e^(1/x)]/x^2 dx =
= Integral de [e^(1/x)] 1/x^2 dx =
Sustitución:
u = 1/x
du = -1/x^2 dx ==> -du = 1/x^2 dx
= Integral de (e^u)(-du) =
= - Integral de e^u du =
= -e^u =
= - e^(1/x) + C
Please integral of e^x^3/x^3
Hi! It is a non-elementary integral...
what if e^(2/x)/x^2 ?
Hi! It doesn't change a lot. Let's see:
Integral of e^(2/x)/x^2 dx =
= Integral of e^(2/x) (1/x^2) dx =
Substitution:
u = 2/x = 2x^(-1)
du = -2x^(-2) dx = -2/x^2 dx ==> du/-2 = 1/x^2 dx
= Integral of e^u du/-2 =
= (-1/2)Integral of e^u du =
= (-1/2)e^u =
= (-1/2)e^(2/x) + C
Hope it helped! ;-D
Integrals ForYou thank you so much :D
You're welcome! 😉👍
integration of (e^(-1/x)+1)/x^(2)
Hi, here you have the solution:
Integral of (e^(-1/x)+1)/x^2 dx =
= Integral of ( e^(-1/x)/x^2 + 1/x^2 ) dx =
= Integral of e^(-1/x) (1/x^2)dx + Integral of 1/x^2 dx =
Substitution:
u = -1/x
du = (1/x^2)dx
= Integral of e^u du + Integral of x^(-2) dx =
= e^u + x^(-2+1)/(-2+1) =
= e^(-1/x) + x^(-1)/(-1) =
= e^(-1/x) - 1/x + C
@@IntegralsForYou THANK YOU SIR
Himanshu Soni You're welcome! 💪
Could you help me with integration of dx/(e^x-1)^2 ?
Of course! Here you have the solution:
Integral of 1/(e^x-1)^2 dx =
Substitution:
u = e^x-1 ==> u+1 = e^x
du = (e^x)dx = (u+1)dx ==> du/(u+1) = dx
= Integral of 1/u^2 du/(u+1) =
= Integral of 1/(u^2)(u+1) du =
= ua-cam.com/video/GyL5qvNe4bg/v-deo.html =
= -ln|u| - 1/u + ln|u+1| =
= -ln|e^x-1| - 1/(e^x-1) + ln|e^x-1+1| =
= -ln|e^x-1| - 1/(e^x-1) + ln|e^x| =
= -ln|e^x-1| - 1/(e^x-1) + x + C
Hope it helped! ;-D
@@IntegralsForYou could you make a video on this one? I am confused
@Mustafa Attarwala I have a TO DO list for the videos, I can add this one to the list but you will have to wait a lot of time for the video to be uploaded. I suggest you to ask here in a comment the doubts you have. If you want to send me a picture you can do it on instagram 😉
@@IntegralsForYou Thanks so much! How to split into partial fractions, if one of the function is a square? How do we get the values after splitting?
@@mustafaattarwala7085
If we have , example, 1/((x-a)^3(x-b)(cx^2+d)) then:
1/((x-a)^3(x-b)(cx^2+d)) = A/(x-a) + B/(x-a)^2 + C/(x-a)^3 + D/(x-b) + (Ex+F)/(cx^2+d) = ...
I suggest you to read the examples section of wikipedia:
en.wikipedia.org/wiki/Partial_fraction_decomposition
Also an example to get the values: If we have the integral of (3x^2+7x+5)/(x-2)(x-3)(x-4) then:
(3x^2+7x+5)/(x-2)(x-3)(x-4) = A/(x-2) + B/(x-3) + C/(x-4) =
= A(x-3)(x-4)/(x-2)(x-3)(x-4) + B(x-2)(x-4)/(x-3)(x-1)(x-2) + C(x-2)(x-3)/(x-4)(x-2)(x-3) =
= [ A(x-3)(x-4) + B(x-2)(x-4) + C(x-2)(x-3) ]/(x-2)(x-3)(x-4)
Now we have two options:
1.
We have (3x^2+7x+5)/(x-2)(x-3)(x-4) = [ A(x-3)(x-4) + B(x-2)(x-4) + C(x-2)(x-3) ]/(x-2)(x-3)(x-4) then:
If x=2 ==>
3*2^2 + 7*2 + 5 = A(2-3)(2-4) + B(2-2)(2-4) + C(2-2)(2-3)
12 + 14 + 5 = A(-1)(-2) + B(0)(-2) + C(0)(-1)
31 = 2A ==> A = 31/2
If x=3 ==>
3*3^2 + 7*3 + 5 = A(3-3)(3-4) + B(3-2)(3-4) + C(3-2)(3-3)
27 + 21 + 5 = A(0)(-1) + B(1)(-1) + C(1)(0)
53 = -B ==> B = -53
If x=4 ==>
3*4^2 + 7*4 + 5 = A(4-3)(4-4) + B(4-2)(4-4) + C(4-2)(4-3)
48 + 28 + 5 = A(1)(0) + B(2)(0) + C(2)(1)
81 = 2C ==> C = 81/2
Then:
A = 2/21
B = -53
C = 81/2
2.
(3x^2+7x+5)/(x-2)(x-3)(x-4) =
= [ A(x-3)(x-4) + B(x-2)(x-4) + C(x-2)(x-3) ]/(x-2)(x-3)(x-4) =
= [ A(x^2-7x+12) + B(x^2-6x+8) + C(x^2-5x+6) ]/(x-2)(x-3)(x-4) =
= [ Ax^2 - 7Ax + 12A + Bx^2 - 6Bx + 8B + Cx^2 - 5Cx + 6C ]/(x-2)(x-3)(x-4) =
= [ (A+B+C)x^2 + (-7A-6B-5C)x + (12A+8B+6C) ]/(x-2)(x-3)(x-4)
==>
3x^2+7x+5 = [ (A+B+C)x^2 + (-7A-6B-5C)x + (12A+8B+6C) ]/(x-2)(x-3)(x-4)
==>
3 = A + B + C ==> C = 3 - A - B
7 = -7A - 6B - 5C ==> 7 = -7A - 6B - 5(3-A-B) ==> 7 = -7A - 6B - 15 + 5A + 5B ==> B = -2A - 22
5 = 12A + 8B + 6C ==> 5 = 12A + 8(-2A-22) + 6(3-A-(-2A-22)) ==> 5 = 12A - 16A - 176 + 18 - 6A + 12A + 132 ==> 5 + 176 - 18 - 132 = 12A - 16A - 6A + 12A ==> 31 = 2A ==> A = 31/2
==>
A = 31/2
B = -2A - 22 = -2(31/2) - 22 = -31 - 22 = -53
C = 3 - A - B = 3 - 31/2 - (-53) = 6/2 - 31/2 + 106/2 = 81/2
Then:
A = 31/2
B = -53
C = 81/2
Hope this answered helped! :-D I have a list of this kind of integrals where I mostly use the second way to find A,B,C...:
Integration by partial fraction decomposition: ua-cam.com/play/PLpfQkODxXi4-9Ts0IMGxzI5ssWNBT9aXJ.html
hi please can you solve this x/e^2x dx with limit of upper 1nad lower 0
Hi! here you have the solution:
Integral of x/e^(2x) dx =
= Integral of x*e^(2x) dx =
Parts: Integral of u dv = uv - Integral of v du
u = x ==> du = dx
dv = e^(-2x) dx ==> v = (-1/2)e^(-2x)
= x*(-1/2)e^(-2x) - Integral of (-1/2)e^(-2x) dx =
= x*(-1/2)e^(-2x) + (1/2)Integral of e^(-2x) dx =
= x*(-1/2)e^(-2x) + (1/2)(-1/2)e^(-2x) =
= x*(-1/2)e^(-2x) - (1/4)e^(-2x) =
= (-1/4)(2x+1)e^(-2x) + C
Then:
Integral of x/e^(2x) dx from 0 to 1 =
= (-1/4)(2*1+1)e^(-2*1) - (-1/4)(2*0+1)e^(-2*0) =
= (-1/4)(3)e^(-2) - (-1/4)(1)e^0 =
= (-3/4)e^(-2) + 1/4 =
= 0.1484...
@@IntegralsForYou omg thank you Sir you saved my life
ISAB Channel 😊😊Anytime! 👍👍
Tnx
You're welcome! ☺
Porque le pone - a la integral y al -1/x2?
Hola!
Como la derivada de 1/x es -1/x^2 y sólo tenemos 1/x^2 , podemos multiplicar por -1 dentro y fuera de la integral, o lo que es lo mismo, poner el signo "-" dentro y fuera de la integral. Así ya tenemos -1/x^2 que es lo que queríamos.
Si lo ves más fácil te dejo otra manera de aplicar el método de sustitución para esta integral:
Integral de e^(1/x)/x^2 dx =
= Integral de e^(1/x) 1/x^2 dx =
Sustitución:
u = 1/x
du = -1/x^2 dx ==> -du = 1/x^2 dx
= Integral de e^u (-du) =
= - Integral de e^u du =
= - e^u =
= - e^(1/x) + C
Hay mucha gente que lo entiende mejor así. Personalmente los dos métodos me parecen igual de correctos, así que escoge! ;-d
jajajajajaja no mams
Y si e estubiera elevado al lnx como queda
Puedes escribirme la expresión? Porque por lo que entiendo quieres hacer e^(ln(x)) y eso es igual a "x".
thankssss
Hi Abel! I've seen you were asking for this video and you have already found it! 😉I share the link in the other comment in case somebody wants to know it! Hope my channel is helping you! 👍
@@IntegralsForYou Thank you very much and if I am going to share it, I also wanted to know if you could make me the integral (2x-4) cosx (x ^ 2-4x) dx
@@asterixmusic4928 Of course, here you have:
Integral of (2x-4)cos(x^2 - 4x) dx =
= Integral of cos(x^2 - 4x) (2x-4)dx =
Substitution:
u = x^2 - 4x
du = (2x-4)dx
= Integral of cos(u) du =
= sin(u) =
= sin(x^2 - 4x) + C
;-D
@@IntegralsForYou sorry i forgot to put the x next to the cos it would be cosx, but would it still be the same?
@@asterixmusic4928 Sorry, Abel, I don't understand your question...
de donde sale el -1 del du?
Hola marlyz johana burgos barrera! El -1 sale de la formula de derivación de x^n:
Derivada de x^n = n*x^(n-1)
Derivada de x^(-1) = (-1)x^(-1-1) = -x^(-2) = -1/x^2
Un saludo!
kyc vieja lesbiana
cool
Thanks! 😊
And if: e^1/x + 2/ x^2 dx?
Hi! I don't understand which expression you are asking for:
Integral of e^(1/x + 2/x^2) dx ?
Integral of e^( 1/(x + 2) + 1/x^2 ) dx ?
Integral of ( e^(1/x) + 2/x^2 ) dx ?
Well, that was easy
It was easy for you 😉
Lol, my book says answer is -(e^1/x +1/x) +c. :O
😜
768.
😉
isn't it u = 1/x^2???
Hi suhreeen, I don't think so... If you let u=1/x^2 then you will have to do a second substitution t=sqrt(u), but it is easier to do t=1/x directly [ t=sqrt(u)=sqrt(1/x^2)=1/x ]:
Integral of e^(1/x) 1/x^2 dx =
Substitution:
u = 1/x^2 = x^-2 ==> sqrt(u) = 1/x
du = -2x^-3 dx = -2/x^3 dx ==> x/-2 du = 1/x^2 dx ==> sqrt(u)/-2 du = 1/x^2 dx
= Integral of e^sqrt(u) sqrt(u)/-2 du =
= (-1/2)Integral of e^sqrt(u) sqrt(u) du =
Substitution:
t=sqrt(u)
dt = 1/2sqrt(u) du ==> 2 dt = sqrt(u) du
= (-1/2)Integral of e^t 2 dt =
= (-2/2)Integral of e^t dt =
= - e^t =
= - e^sqrt(u) =
= - e^(1/x) + C
Hope I answered your question. If not, ask me again with more details ;-D
助かったやで サンガツ
You're welcome! ;-D
Pls help
Hi! Here you have the solution:
Integral of e^(1/x)/x^3 dx =
= Integral of (e^(1/x))*(1/x)*(1/x^2)*dx =
Substitution:
u = 1/x
du = (-1/x^2)*dx ==> -du = (1/x^2)*dx
= Integral of (e^u)*u*(-du) =
= - Integral of u*e^u du =
= - ua-cam.com/video/oOlGshv2ycc/v-deo.html =
= - ( u*e^u - e^u ) =
= (e^u)(1 - u) =
= (e^(1/x))(1 - 1/x) + C
Hope it helped! ;-D
Thanks
You're welcome!
Thanks
You're welcome! ;-D