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Пікірлер
@samarthmn6766 Күн бұрын
You can just split the x^5 into x^4 and x now assume x^2 as t and u have the derivative of x^2 in numerator along with t^2 above and 1+t in denominator after that its basic integration quicker method
@IntegralsForYou Күн бұрын
Tooootally agree! Let me write the steps in case anyone wants to know more: Integral of x^5/(1+x^2) dx = = Integral of (x^4)*x/(1+x^2) dx = = Integral of (x^2)^2/(1+x^2) x*dx = Substitution: t = x^2 dt = 2x*dx ==> dt/2 = x*dx = Integral of t^2/(1+t) dt/2 = = (1/2)*Integral of t^2/(1+t) dt = Substitution: u = 1+t ==> u-1 = t du = dt = (1/2)*Integral of (u-1)^2/u du = = (1/2)*Integral of (u^2-2u+1)/u du = = (1/2)*Integral of (u^2/u - 2u/u + 1/u) du = = (1/2)*Integral of (u - 2 + 1/u) du = = (1/2)*(u^2/2 - 2u + ln|u|) = = (1/4)*u^2 - u + (1/2)*ln|u| = = (1/4)*(1+t)^2 - (1+t) + (1/2)*ln|1+t| = = (1/4)*(1+2t+t^2) - 1 - t + (1/2)*ln|1+t| = = 1/4 + (1/2)*t + (1/4)*t^2 - 1 - t + (1/2)*ln|1+t| = = -3/4 - (1/2)*t + (1/4)*t^2 + (1/2)*ln|1+t| = = -3/4 - (1/2)*x^2 + (1/4)*x^4 + (1/2)*ln|1+x^2| + C = = (1/4)*x^4 - (1/2)*x^2 + (1/2)*ln|1+x^2| + C' (where C' = -3/4+C is also a constant) Thanks for your comment! 💪
@FunkyPhilMusic 2 күн бұрын
why can’t you multiply times 1/2 so you have radx’s derivate in the integral and then proceed
@IntegralsForYou Күн бұрын
Hi! The derivative of sqrt(x) is 1/2sqrt(x) and we don't have 1/sqrt(x) in the expression. We can do it in the case of the integral of sin(sqrt(x))/sqrt(x). I did the video here: kzbin.info/www/bejne/iIW7loKcjZdrpZo Hope it helped! 💪
@xasanbayev_olimjon_007 2 күн бұрын
Anyone 2025 1:38
@malluboytech9426 6 күн бұрын
nice video sir
@IntegralsForYou 4 күн бұрын
Thank you! ❤
@sabrina-tq6mc 7 күн бұрын
hi can you pls integrate (x-1)/(x+2) dx
@IntegralsForYou 7 күн бұрын
Of course! Here you have the solution: Integral of (x-1)/(x+2) dx = = Integral of (x-1+2-2)/(x+2) dx = = Integral of (x+2-2-1)/(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! 💪
@sabrina-tq6mc 7 күн бұрын
@IntegralsForYou thankyou
@IntegralsForYou 4 күн бұрын
My pleasure! ❤
@KenKaneki-ol1wx 8 күн бұрын
You dumb cant you open your damn mouth and speak and explain us
@IntegralsForYou 7 күн бұрын
Hi! As I say in my website ( integralsforyou.com/frequently-asked-questions#why-dont-videos-have-sound ), in this channel I do integrals without sound. If you have any doubts about this particular integral you can ask in a comment and I'll try to do my best when answering 😉 I hope you will learn and practice a lot in this channel! 💪
@SV-qr1km 9 күн бұрын
Thank you
@IntegralsForYou 8 күн бұрын
My pleasure! ❤
@fatslots8006 12 күн бұрын
Isnt A-C=-1?
@IntegralsForYou 7 күн бұрын
Hi! We have to take the 1 of the numerator and not the -1 from the denominator: 1/(x^3-1) = [(A+B)x^2 + (A-B+C)x + (A-C)]/(x^3-1) 1 = (A+B)x^2 + (A-B+C)x + (A-C) 0x^2 + 0x + 1 = (A+B)x^2 + (A-B+C)x + (A-C) Then: 0 = A + B 0 = A - B + C 1 = A - C Hope it helped! 💪
@parhamgh7481 12 күн бұрын
Thank you
@IntegralsForYou 8 күн бұрын
My pleasure! ❤
@Phonkist25 13 күн бұрын
Nim avvan bos########*****
@Linspecteurduparanormal 13 күн бұрын
Hello, your video is great but i have a question, i use an another technic to solve this integral, i say in the beginning that ln(1-x2) = ln(-(x+1)(x-1)), after i separate the ln with the rule of addition, and decompose the integral in two integral : integral of -ln(x+1) and integral of -ln(x-1). I put u=ln(x+1) and v = ln(x-1) (variable changement in the integrals) and obtain integer of ue^^u du + integer of ve^^v. Finally to solve that, i use the method of integration by part, but i don't find the same result i believe, can you say at me if someting is wrong in my raisonnement ? Thank you very much (I hope my english is correct ahah)
@IntegralsForYou 8 күн бұрын
Hi! I'm sorry but I forgot the -2x in the solution in the video... Let's see the final answer using your method: Integral of ln(1-x^2) dx = = Integral of ln((1+x)(1-x)) dx = = Integral of [ln(1+x) + ln(1-x)] dx = = Integral of ln(1+x) dx + Integral of ln(1-x) dx = Substitution: u = 1+x du = dx Substitution: v = 1-x dv = -dx ==> -dv = dx = Integral of ln(u) du + Integral of ln(v) (-dv) = = Integral of ln(u) du - Integral of ln(v) dv = We know the integral of ln(x) is x*(ln(x)-1) then: = u*(ln(u)-1) - v*(ln(v)-1) = = (1+x)*(ln(1+x)-1) - (1-x)*(ln(1-x)-1) = = ln(1+x) - 1 + x*ln(1+x) - x - ln(1-x) + 1 + x*ln(1-x) - x = = x*[ln(1+x)+ln(1-x)] + ln(1+x) - ln(1-x) - 2x = = x*ln((1+x)(1-x)) + ln(1+x) - ln(1-x) - 2x = = x*ln(1-x^2) + ln(1+x) - ln(1-x) - 2x + C
@MikeJenson 15 күн бұрын
Hi there, thank you so much for your service. I just discovered your channel - 9 YEARS MAN! Das crazy. There are things that need to exist in this world, and this is one of them. I have often fanaticised about doing something similar myself when I get really good at integration - but I'm so glad you have. I can sleep easy! Thank you from struggling maths students everywhere.
@IntegralsForYou 15 күн бұрын
Thank you so much for your comment! 9 years already... time flies hehe Thank you for your words, they keep me motivated to continue this project to help lots of students! If you could share this channel to your friends or people who may need it, it would be amazing! Thanks again! ❤❤
@Factz_Hub_ 19 күн бұрын
Totally wrong 😢😢
@IntegralsForYou 15 күн бұрын
Hi! It is possible! May I know where I did it wrong? Thanks in advance!
@AmrKhaled-i8y 19 күн бұрын
ايه العبط ده 😂
@AmrKhaled-i8y 19 күн бұрын
ايه العبط ده 😂
@AmrKhaled-i8y 19 күн бұрын
حضرتك أحول
@amittomar419 20 күн бұрын
hi sir can you uplode one shot forintegration for jee mains and advanced please
@Ashishbhartiii 21 күн бұрын
Bro use half angle formula from trigonometry 1+cosx=2cos^2(x/2)
@IntegralsForYou 15 күн бұрын
Yes! You are right! There is more than one method to solve this integral!
@alishipuden6224 22 күн бұрын
i didnt study integration yet but i still got the same formula, i only used known primitives, we have f(x)=1/cos^6 =(tan^2 +1)^2 *tan' =tan^4 *tan' +tan^2 *tan' +tan' and this is equal to the same thing in the video. thats for the video i really wanted to check if my answer was right
@IntegralsForYou 15 күн бұрын
Nice! Glad you found my video to check your solution! ❤
@awinbmx7418 22 күн бұрын
In high school I got this problem on a exam, and substitution was not included in our course, teacher said there was another way to solve it, I sat for 1 hour with different trig formulas and identities and it was impossible, was my teacher mentally unstable?
@IntegralsForYou 15 күн бұрын
Hi! I don't know... did your teacher say something more about this integral?
@awinbmx7418 15 күн бұрын
@@IntegralsForYou in our course, we had product rule for derivative, and quotient rule, and the trig entities derived from the unit circle, with this info we were supossed to solve this integral
@awinbmx7418 15 күн бұрын
The question was supossed to give 4 , A-grade marks on the exam, (tough question)
@awinbmx7418 15 күн бұрын
@@IntegralsForYou is there a way to substitute without variable substitution? Like in a high school level
@awinbmx7418 15 күн бұрын
@@IntegralsForYou can you try solve it with product rules and trig identities
@DanDCool 22 күн бұрын
thanks so much sir i was pulling some x = secT its not even funny👹
@IntegralsForYou 15 күн бұрын
My pleasure! x=sec(t) is a nice try but I think it is better the method used in the video...
@ВладимирРостиков 23 күн бұрын
Динаху
@qopw 23 күн бұрын
you can also differentiate under the integrand here and skip substituting. pretty cool
@AmnNino 26 күн бұрын
🤍🤍💯
@IntegralsForYou 26 күн бұрын
❤
@montblanc2.033 26 күн бұрын
You can also multiply both numerator and denominator by 1-cos(x), obtaining the integral of (1-cos(x))/sen²(x), which is cosec²(x) - [cos(x)/sen²(x)]. The integral of cosec²(x) is -cotg(x)+C, and the integral of cos(x)/sen²(x): u=sen(x), du=cos(x)dx, so we get the integral of 1/u², and that is -1/u [-1/sen(x)] +C. So the answer is -cotg(x) + cosec(x) + C; C € R
@IntegralsForYou 26 күн бұрын
Hi! Totally agree! I did the video using the method you proposed here: kzbin.info/www/bejne/mWimYYJsoLeLnbc 💪
@IntegralsForYou 26 күн бұрын
List of solved integrals in a PDF file in my Patreon page: www.patreon.com/integralsforyou/shop ❤
@Akash902 29 күн бұрын
Integration of e^x^2 dx ?
@IntegralsForYou 29 күн бұрын
Hi! The integral of e^(x^2) dx is a non-elementary integral, it cannot be expressed in terms of finite standard functions...
@maxtv1728 29 күн бұрын
I've been trying to solve that integral for an hour, after 3 hours of solving others. My brain was too fried for this haha
@IntegralsForYou 29 күн бұрын
haha I think you should maybe take some pauses while integrating... 🙃
@theJune666 29 күн бұрын
Святой человек, спасибо, что ты есть Если б не ты, то меня бы съела математичка Выходи за меня 👉👈💗💗💗💗💗💗💗
@IntegralsForYou 28 күн бұрын
🤣❤
@AbdelhayFerqass Ай бұрын
oh man you're a legend!
@IntegralsForYou Ай бұрын
Thank you! This is a nice integral to solve! 💪
@rupamsharma8776 Ай бұрын
Omg.😱😱😱
@IntegralsForYou Ай бұрын
😱😱
@AnkitRaj-fh8zw Ай бұрын
Bolkar bnaiye bhaiya 😢😢❤❤
@puspayadav6503 Ай бұрын
Thankyou 💕😇
@IntegralsForYou Ай бұрын
You’re welcome! 😊
@luffyblood1961 Ай бұрын
Bro idonot know your alien or else but you wrting style is crzy
@IntegralsForYou Ай бұрын
I am an alien... 🤫
@HarshRaj-bd6og Ай бұрын
Khaile naikh ka
@lovlday6847 Ай бұрын
thanks men 👍
@IntegralsForYou Ай бұрын
My pleasure! ❤
@lovesandhu_047 Ай бұрын
Yr thankyou thankyou so much
@IntegralsForYou Ай бұрын
Most welcome! 😊
@RaushanKumar-ip2li Ай бұрын
Bhai itna me to ala Bala pe chale jayenge However hard work And lack of smart work
@MyLegsAreKindaLong Ай бұрын
bro alway have answer for even the most random equation
@IntegralsForYou Ай бұрын
🤭
@aaryanjha83 Ай бұрын
Love u ,I am from India and I have seen so many questions from u ,thank u
@IntegralsForYou Ай бұрын
Thank you for watching! Glad it helped! ❤
@lukasarpi6851 Ай бұрын
Thanks, I really helped me 🫶
@IntegralsForYou Ай бұрын
Glad it helped! ❤
@msreddevil897 Ай бұрын
Thank you for your teaching sir
@IntegralsForYou Ай бұрын
My pleasure! ❤
@kushwanshi_99 Ай бұрын
Thanks a lot bhaiya . This formulas are very useful. ❤
@IntegralsForYou Ай бұрын
Glad to hear that! ❤
@kyumi133 Ай бұрын
is it the only way to resolve it? just being curious
@IntegralsForYou Ай бұрын
Hi! I think it is the only way... you can try doing u=sqrt(1+x^2) but you will end up doing a trig substitution...
@jorgemedeiros6929 Ай бұрын
This video helped me a lot! Thank you very much!
@IntegralsForYou Ай бұрын
Glad it helped! ❤❤
@harshitsharma4697 Ай бұрын
thank you
@IntegralsForYou Ай бұрын
You're welcome! ❤❤
@borasahin5068 Ай бұрын
BUNUNLA NE YAPILIYOR HOCAM
@borasahin5068 Ай бұрын
WOW AMAZİNG
@IntegralsForYou Ай бұрын
Thank you! ❤
@weird1209 Ай бұрын
But isn't integration of 2x , x^2 then why have we written it as 1/2
@IntegralsForYou Ай бұрын
Hi! The integral of 2x is x^2 but in this case we have to derive it. When we do u=2x we must derive in both sides of the equality and add "du" and "dx". Then , we have to isolate "dx" in order to proceed to the substitution.
@mamadetaslimtorabally7363 Ай бұрын
Great as usual. Integration is a beautiful chapter and for this question here the BIC pen is just the ideal pen for it.
@IntegralsForYou Ай бұрын
It really is! Thank you! ❤❤