
English: 
- [Voiceover] Try to evaluate
the following integral.
So assuming you've had a go at it,
so let's work through this together.
And if at any point you get inspired,
always feel free to pause the video
and continue on with it on your own.
So the first thing that
might have jumped out at you
we have a rational expression.
The degree in the numerator is the same as
the degree in the nominator,
so maybe a little bit
of algebraic long division is called for.
So let's do that.
Let's take X-squared minus one
and divide it into X-squared,
put that in a different color,
divide it into X-squared
plus X minus five.
So X-squared plus X minus five.
And so let's look at
the highest degree terms
how many times does
X-squared go into X-squared?
It goes one time, let me
write this in a new color.
Goes one time.
One times X-squared minus
one is just going to be
X-squared minus one.
Now you subtract this
green expression from

Portuguese: 
﻿Tente avaliar
a seguinte integral.
Assim, supondo
que você tentou,
vamos trabalhar juntos.
E se você se inspirar,
sinta-se livre para
pausar o vídeo
e continuar com isto
por conta própria.
A primeira coisa que
deve ter notado,
temos uma
expressão racional.
O grau no numerador
é o mesmo que o
grau no denominador,
então talvez
algébra de divisão
longa é possível.
Vamos fazer isto.
Vamos pegar x ao menos um
e dividi-lo em x ao quadrado-
colocarei em uma cor diferente-
dividi-lo em x ao quadrado
mais x menos cinco.
X ao quadrado
mais x menos cinco.
Vamos olhar para os
termos de maior grau
Quantas vezes x ao quadrado
cabe em x ao quadrado?
Uma vez, deixe-me escrever
isto em uma nova cor.
Uma vez.
Um vezes x ao quadrado
menos um será
x ao quadrado menos um.
Agora você subtrai esta
expressão verde por

Czech: 
Zkuste vypočítat následující integrál.
Nejspíš jste to už zkusili, takže
teď se na to podíváme společně.
A pokud vám někde napovím,
klidně zase zastavte
video a pokračujte sami.
První, co vás asi napadne, je,
že tu máme racionální výraz.
Stupeň v čitateli je stejný
jako stupeň ve jmenovateli,
takže to asi bude chtít nějaké dělení.
Tak se do toho dejme.
Vezmeme (x na druhou) minus 1
a vydělíme tím ((x na druhou)…
Udělám to jinou barvou.
Vydělíme tím
((x na druhou) plus x minus 5).
Takže ((x na druhou) plus x minus 5).
Podívejme se na členy nejvyššího stupně.
Kolikrát se (x na druhou)
vejde do (x na druhou)?
Vejde se to tam jednou,
udělám to jinou barvou.
Jednou.
1 krát ((x na druhou) minus 1) bude
prostě (x na druhou) minus 1.
A teď odečtete tenhle zelený
výraz od toho světle fialového

Thai: 
ลองหาค่าอินทิกรัลต่อไปนี้กัน
สมมุติว่าคุณได้ลองแล้ว
ลองทำอันนี้ไปด้วยกันดู
และถ้าคุณเกิดอยากลองเองตอนไหน
ก็หยุดวิดีโอนี้
และทำต่อด้วยตัวเองได้
อย่างแรกที่อาจสะดุดตาคุณ
คือว่าเรามีพจน์ตรรกยะ
ดีกรีของตัวเศษเท่ากับ
ดีกรีของตัวส่วน บางที
อาจต้องใช้การหารยาว
ลองทำดู
ลองนำ x กำลังสองลบ 1
มาหารไปใน x กำลังสอง ใส่อีกสีนะ
เอามันไปหาร x กำลังสองบวก x ลบ 5
x กำลังสองบวก x ลบ 5
แล้วลองดูเทอมดีกรีสูงสุด
x กำลังสองไปหาร x กำลังสองได้กี่ครั้ง?
มันได้ 1 ครั้ง ขอผมเขียนด้วยสีใหม่นะ
ได้ 1 ครั้ง
1 คูณ x กำลังสองลบ 1 จะเท่ากับ
x กำลังสองลบ 1
ทีนี้ คุณลบพจน์สีเขียวนี้

Bulgarian: 
Опитай да изчислиш
следния интеграл.
Предполагам, че опита,
а сега да го направим заедно.
Ако в някакъв момент почувстваш
прилив на вдъхновение,
винаги можеш да спреш видеото
на пауза
и да продължиш самостоятелно.
Първото нещо, което може би 
ти хрумва, е, че това е рационален израз.
Степенният показател в
числителя е същия като
степента в знаменателя, така че
можем да преработим алгебрично.
Да го направим.
Да вземем (х^2 –1)
и да го разделим на х^2...
ще използвам различен цвят –
да разделим на него 
 х^2 + х – 5.
Да видим членовете с
най-висока степен,
колко пъти се съдържа
х^2 в х^2?
Съдържа се един път, като
ще използвам нов цвят.
Съдържа се един път.
Едно по (х^2 – 1) ще бъде
просто (х^2 – 1).
Сега изваждаме този
зелен израз от

Korean: 
이 적분을 계산해봅시다
당신이 해봤다고 가정하고
이 문제를 함께 풉시다
그리고 만약 어떤 부분에서 영감을 얻으면
항상 동영상을 자유롭게 멈추세요
그리고 문제를 혼자서 계속해 보세요
그래서 첫번째로 나타난것은
우리가 유리식을 가지고 있다는 것입니다
분자에 있는 degree가 nominator의 degree와
같고, 아마 약간의
대수학의 장제법이 요구된다
이제 해봅시다
x제곱 빼기1
그리고 그것을 x제곱으로 나눕시다 이것을 다른색으로 놓을게요
이것을 x제곱 더하기 x빼기 5로 나눕시다
그래서 x제곱 더하기 x빼기 5
그
x제곱이 x제곱에 몇번가있나요

Korean: 
그리고 명백히 이것은 꽤 직설적이다

Portuguese: 
esta expressão roxa
ou eu poderia adicionar
o negativo nele,
deixe-me tomar
o negativo dele.
Será x ao quadrado
negativo mais um
e obteremos os x's ao quadrado -
x quadrado menos
x quadrado é zero,
então eles se cancelam
e ficaremos com
x e o cinco negativo mais
um é quatro negativo.
Portanto, temos x menos
quatro sobrando.
Podemos reescrever a
expressão no qual tentamos
encontrar a antiderivada.
Podemos reescrever como
um mais x menos quatro
sobre x ao quadrado
menos um.
Talvez faça na cor roxa
pois já usei isto em roxo.
sobre x ao quadrado menos um.
Então fizemos uma coisa, agora
temos um menor grau
no numerador do que
temos no denominador.
Obviamente, isto é
bem simples,
tomar a antiderivada,

Czech: 
nebo můžu říct, že přičítám
jeho záporný násobek.
Takže to bude (-x na druhou) plus 1
a teď se podíváme na ta (x na druhou).
x na druhou minus x na druhou je 0,
takže tyhle se zruší a zůstane
nám x a -5 plus 1, což je -4.
Takže máme zbytek x minus 4.
Takže můžeme přepsat ten výraz,
jehož primitivní funkci chceme najít.
Můžeme to přepsat jako 1 plus ((x minus 4)
děleno ((x na druhou) minus 1)).
Asi to udělám fialově, když
už jsem to tak dělal předtím.
…děleno ((x na druhou) minus 1).
Takže teď jsme udělali tohle,
teď máme v čitateli nižší
stupeň než ve jmenovateli.
A k tomuhle je samozřejmě
lehké najít primitivní funkci,

Bulgarian: 
този светло розов израз
или просто го добавяме
със знак минус, така че
ще взема този минус от тук.
Това ще стане –х^2 + 1,
и после тези х^2 се унищожават,
х^2 минус x^2 е нула,
тези се унищожават
и ни остава само
х и –5 плюс 1 е равно
на –4.
Значи ни остава х – 4.
Можем да преработим израза,
така че да опитаме
да намерим примитивната 
му функция.
Можем да го преработим като
1 + х – 4
върху (х^2 – 1).
Може би да използвам
лилав цвята,
тъй като вече съм използвал
лилаво.
Върху (х^2 – 1).
Направихме това, че сега
имаме по-ниска степен
в числителя, отколкото
в знаменателя.
Очевидно за този израз
е много по-лесно
да намерим примитивната
функция.

Thai: 
ออกจากพจน์สีชมพู หรือผมบวก
ค่าลบของมัน ขอผม
ใส่ลบของมันนะ
มันจะได้ลบ x กำลังสองบวก 1
และเราจะได้ x กำลังสอง
x กำลังสองลบ x กำลังสองเป็น 0
พวกมันหักล้างกัน แล้วเราจะเหลือ
x กับลบ 5 บวก 1 เป็นลบ 4
เราจึงได้ x ลบ 4 เหลืออยู่
เราก็เขียนพจน์ที่เราพยายาม
หาปฏิยานุพันธ์ใหม่ได้
เราเขียนมันใหม่ได้เป็น 1 บวก x ลบ 4
ส่วน x กำลังสองลบ 1
บางทีผมจะใช้สีม่วงนั้น
เพราะเราใช้มันเป็นสีม่วงไปแล้ว
ส่วน x กำลังสองลบ 1
เราทำไปอย่างหนึ่ง ตอนนี้เรามีดีกรีในตัวเศษ
น้อยกว่าที่เรามีในตัวส่วนแล้ว
แน่นอน อันนี้ตรงไปตรงมา
เวลาหาปฏิยานุพันธ์

English: 
this mauve expression or I could just add
the negative on it, so let me just take
the negative of it.
So it's going to be
negative X-squared plus one
and we're going to get the X-squared's
X-square minus X-square is zero,
so those cancel out and
we're going to be left with
X and the negative five
plus one is negative four.
So we have X minus four left over.
So we can rewrite the
expression that we're trying to
find the antiderivative of.
We can rewrite it as one plus X minus four
over X-squared minus one.
Maybe I'll do that in that purple color
since I already used it as the purple.
Over X-squared minus one.
So we did one thing, now
we have a lower degree
in the numerator than we
have in the denominator.
And obviously this is
fairly straight forward,
take the antiderivative of,

Thai: 
แต่เราทำอะไรได้ตอนนี้?
มันไม่ชัดนักถ้าเราดู x กำลังสองลบ 1
อนุพันธ์ของมันเป็น 2x
ซึ่งมีดีกรีเท่ากับตัวนี้
แต่มันไม่ใช่ x ลบ 4 มันจึงดูเหมือนว่า
การแทนที่ u จะไม่ช่วยอะไรเรานัก
แล้วเราทำอะไรได้?
เราใช้เครื่องมืออีกอย่าง
ในพีชคณิตได้ เราจะใช้
การกระจายเศษส่วนย่อย
ซึ่งก็คือการเขียนตัวนี้เป็นผลบวก
ของพจน์ตรรกยะสองตัวที่
มีดีกรีในตัวส่วนน้อยลง
ผมหมายความว่าอะไร?
เทอมนี่ตรงนี้
x ลบ 4 ส่วน x กำลังสองลบ 1
เราเขียนมันใหม่ได้เป็น
x ลบ 4 ส่วน, แทนที่จะเป็น x กำลังสองลบ 1
เราแยกอันนี้ได้
นี่คือ x บวก 1 คูณ x ลบ 1
ลองเขียนมันดู
นี่คือ x บวก 1 คูณ x ลบ 1
เวลาเราคิดถึงการกระจายเศษส่วนย่อย
เราบอกว่า โอเค เราเขียนพจน์นี้เป็นผลบวก

Korean: 
그러나 지금 우리가 무었을 알고있나

Czech: 
ale co uděláme teď?
Není to jasné.
Když se podíváme na (x na druhou)
minus 1, tak derivace bude 2x,
což má stejný stupeň jako tohle,
ale není to x minus 4, takže to nevypadá,
že by nám pomohla substituce.
Takže co budeme dělat?
Teď můžeme vzít další algebraický nástroj,
uděláme rozklad na parciální zlomky.
Je to vlastně přepsání tohoto jako
součet dvou racionálních výrazů
s menším stupněm ve jmenovateli.
Co tím přesně myslím?
Tuhle věc, (x minus 4) děleno
((x na druhou) minus 1),
můžeme přepsat jako (x minus 4) děleno…
(x na druhou) minus 1 můžeme rozložit.
To je (x plus 1) krát (x minus 1).
Takže to napišme, je to
(x plus 1) krát (x minus 1).
Máme-li rozklad na
parciální zlomky, řekneme si:

Bulgarian: 
Но какво да направим сега?
Не е ясно, ако разгледаме
(х^2 – 1)
производната му ще бъде 2х,
което е същата степен като това,
но това не е (х – 4), така че
не изглежда вероятно
да можем да използваме
интегриране със заместване.
Какво можем да направим?
Сега можем да използваме
друг инструмент
от нашия алгебричен инструментариум,
разлагане на елементарни дроби.
Това означава да го запишем
като сбор
от два рационални израза
с по-ниска степен на знаменателя.
Какво имам предвид?
Този член ето тук,
(х – 4) върху (х^2 – 1),
можем да го преработим като
(х – 4) вместо върху (х^2 – 1), 
можем да изнесем пред скоби това.
Това е (х + 1) по (х – 1).
Да запишем това,
(х + 1)(х – 1).
Когато искаме да разложим
елементарни дроби,
можем да представим
това като сума от нещо,

English: 
but what do we do now?
It's not clear if we look
at X-squared minus one
it's derivative would be two-X,
which is the same degree as this,
but it's not X minus four,
so it doesn't look like
you u-substitution it's
going to help us with this.
So what can we do now?
Now we can take out another tool in our
algebraic tool kit, we will do
partial fraction expansion.
Which is essentially
writing this as the sum
of two rational expressions that have
a lower degree in the denominator.
So what do I mean by that?
So this term right over here,
X minus four over X-squared minus one,
we can rewrite that as
X minus four over, instead
of X-squared minus one,
we can factor this.
This is X plus one times X minus one.
So let's write that,
this is X plus one times X minus one.
When we think about
partial fraction expansion
we say okay, can we write this as the sum

Portuguese: 
mas o que fazemos agora?
Não está claro se olharmos
x ao quadrado menos um,
sua derivada seria 2x,
que tem o mesmo grau que isto,
mas não é x menos quatro,
então não parece que
substituição U irá nos
ajudar com isto.
Então o que podemos fazer agora?
Agora podemos usar outra
ferramenta em nosso
kit de ferramenta
algébrica, faremos
expansão em frações parciais.
Que é essencialmente
escrever isto como a soma
de duas expressões racionais que têm
um grau menor no denominador.
O que quero dizer com isso?
Portanto, este termo aqui,
x menos quatro sobre x
quadrado menos um,
podemos reescrever isto como
x menos quatro sobre, em vez
de x quadrado menos um,
podemos fatorar isto.
Isto é x mais um vezes x menos um.
Vamos escrever isto,
isto é x mais um vezes x menos um.
Quando pensamos em
expansão da fração parcial
dizemos OK, podemos
escrever como a soma

English: 
of something, let's call
that A, over X plus one,
plus something else, let's call that B,
plus B over X minus one.
Can we do that?
And to attempt to do
that, if we had just add
these two things, what would we get?
Well we would find a common denominator,
which would be X plus
one times X minus one,
and so you would have, if
any of this looks unfamiliar
I encourage you to review the videos
on partial fraction
expansion, because that's
exactly what we're doing right over here.
But this would be equal to
if you were to add the two
your common denominator
would be the product.
So it would be X plus
one times X minus one.
So the first term I would
multiply the numerator
and the denominator times X minus one.

Bulgarian: 
което ще означа като част А,
върху (х + 1),
плюс нещо друго, което
ще означа като В,
плюс В/(х – 1).
Можем ли да направим това?
За да го направим, трябва
само да съберем
тези две неща, тогава
какво ще получим?
Трябва да намерим
общ знаменател,
който ще бъде (х + 1)(х – 1),
и после трябва...
ако това ти е непознато,
ти препоръчвам да гледаш
видео уроците
за разлагане на елементарни 
дроби, защото това е
точно това, което правим
в момента тук.
Това ще бъде равно –
ако трябва да ти съберем,
общият знаменател ще бъде
равен на тяхното произведение.
Ще бъде (х + 1)(х – 1).
За първия член умножаваме и
числителя,
и знаменателя по (х – 1).

Portuguese: 
de algo, vamos chamar
isto de A, sobre x mais um,
mais alguma outra coisa,
vamos chamar isso de B,
mais B sobre x menos um.
Podemos fazer isso?
Para tentar fazer isto,
se somássemos
estas duas coisas,
o que obteríamos?
Encontraríamos um denominador comum,
que seria x mais
um vezes x menos um,
e assim você teria, se
isto lhe parece estranho,
convido você a rever os vídeos
de expansão em fração
parcial, porque isto
é exatamente o que
estamos fazendo aqui.
Mas isto seria igual
se você adicionasse os dois
seu denominador comum
seria o produto.
Portanto, seria x mais
um vezes x menos um.
O primeiro termo eu
multiplicaria o numerador
e o denominador vezes x menos um.

Thai: 
ของอะไรสักอย่าง 
ลองเรียกมันว่า A ส่วน x บวก 1
บวกอีกอย่าง เรียกมันว่า B
บวก B ส่วน x ลบ 1 ได้ไหม
เราทำได้ไหม?
เวลาทำ ถ้าเราแค่บวก
สองตัวนี้ เราจะได้อะไร?
เราจะหาตัวส่วนร่วม
ซึ่งก็คือ x บวก 1 คูณ x ลบ 1
แล้วคุณจะได้ ถ้าอันนี้ดูไม่คุ้น
ผมแนะนำให้คุณทบทวนวิดีโอ
เรื่องการกระจายเศษส่วนย่อย เพราะมัน
คือสิ่งที่เรากำลังทำตรงนี้เลย
อันนี้จะเท่ากับ ถ้าคุณบวกสองตัวนี้
ตัวส่วนร่วมของคุณจะเป็นผลคูณ
มันจะเป็น x บวก 1 คูณ x ลบ 1
เทอมแรก ผมคูณตัวเศษ
และตัวส่วนด้วย x ลบ 1

Czech: 
"Fajn, můžeme to přepsat jako
součet něčeho, řekněme ,A',
děleno (x plus 1) plus něco jiného,
řekněme ,B', děleno (x minus 1)?"
Můžeme to udělat?
Takže když se o to pokoušíme,
když sečteme tyhle dvě věci, co dostaneme?
Najdeme společného jmenovatele,
což je (x plus 1) krát
(x minus 1), takže máme…
Jestli vám to nepřijde vůbec povědomé,
podívejte se znovu na videa
o rozkladu na parciální zlomky,
protože přesně to tady děláme.
Takže to bude rovno…
Když tohle sčítáme, společným
jmenovatelem bude součin.
Takže to bude (x plus 1) krát (x minus 1).
Takže první člen… Čitatele a
jmenovatele násobím (x minus 1).

Bulgarian: 
Това става А по (х – 1)
плюс В, вторият член,
на който умножаваме
и числителя, и знаменателя
по (х + 1).
И какво получаваме?
Това е равно на Ах...
може би ще взема нов цвят.
Това ще е равно на
Ах – А + Вх + В.
Всичко това е върху ето това,
което преписваме отново.
Всъщност просто ще копирам
и ще поставя това.
Копирам и поставям, мога 
да го използвам отново и отново.
Значи имаме това
върху това.
Да видим, сега можем да групираме 
членовете, съдържащи х.

Thai: 
มันจะเป็น A คูณ x ลบ 1
บวก B, เทอมที่สองผมจะคูณ
ตัวเศษและส่วนด้วย x บวก 1
แล้วเราจะได้อะไร?
อันนี้จะเท่ากับ Ax
บางทีผมจะใช้สีเดียว
อันนี้จะเท่ากับ
Ax ลบ A บวก Bx บวก B
แล้วทั้งหมดนั้นส่วนพจน์นี้ที่เรา
เขียนซ้ำแล้วซ้ำเล่า
ที่จริง ขอผมลอกและวางมันนะ
ลอกและวาง ผมใช้มันหลายครั้งมาก
เรามีอันนั้นส่วนอันนั้น
ลองดู ตอนนี้เราจับกลุ่มเทอม x ได้
เราเขียนอันนี้ใหม่ได้เป็น ถ้าเราคิด Ax บวก Bx

Portuguese: 
Portanto, seria A vezes x menos um
mais B, o segundo
termo eu multiplicaria o
numerador e o denominador
por x mais um.
E então o que obtemos?
Isto será igual Ax,
talvez eu faça isso tufo em uma cor.
Isto será igual a
Ax menos A mais Bx mais B
e depois tudo isso sobre
essas coisas que continuamos
escrevendo.
Na verdade, deixe-me
copiar e colar isso.
Copiar e colar, eu posso usar
aquilo de novo.
Portanto, temos isto sobre aquilo.
Vamos ver, agora podemos
agrupar os termos x's.

English: 
So it would be A times X minus one
plus B, the second term I'll multiply the
numerator and the
denominator times X plus one.
And so what do we get?
This is going to be equal to AX,
maybe I'll do this all in one color.
This is going to be equal to
AX minus A plus BX plus B
and then all of that
over this stuff we keep
writing over and over again.
Actually, let me just copy and paste this.
So copy and paste, I can use
that over and over again.
So we have that over that.
So let's see, now we
can group the X terms.
So we can rewrite this as,
so if we take AX plus BX

Czech: 
Takže to bude ,A' krát (x minus 1) plus B…
u druhého členu násobím
čitatele a jmenovatele (x plus 1).
Takže co dostaneme?
Tohle bude rovno Ax…
Asi to udělám vše jednou barvou.
Tohle bude rovno Ax minus A plus Bx plus B
a to všechno dělíme tímhle,
co tu píšeme celou dobu.
Já to jen zkopíruju a vložím.
Takže kopírovat a vložit, můžu
to používat pořád dokola.
Takže máme tohle děleno tímhle.
Takže se podívejme, teď
dáme dohromady členy s ,x'.

Bulgarian: 
Можем да преработим това, така
че ако вземем Ах и Вх,
това е равно на (А + В) по х.
После имаме –А и В.
Значи +В –А, и просто
ограждам със скоби,
за да групирам тези
константни членове.
Всичко това е делено на –
хубаво е, че копирах това
и го поставих,
върху (х + 1)(х – 1).
И сега следва основният момент
при разлагането на елементарни дроби.
Казваме си: "Добре, 
направихме всичко това,
направихме цялото упражнение,
за да видим дали
имаме някакви А и В, 
за които това е вярно.
Ако има някакви А и В, 
за които това е вярно,
тогава (А + В) трябва да е
коефициентът
на този х-член ето тук.
Значи (А + В) трябва
да е равно на 1,
трябва да е равно на този
коефициент.
(В – А) трябва да е 
равно на константата,
трябва да е равно на –4.

Thai: 
จะเท่ากับ A บวก B คูณ x
แล้วเราจะได้ลบ A กับ B
บวก B ลบ A และผมจะใส่วงเล็บรอบมัน
ผมจับกลุ่มค่าคงที่เหล่านี้ได้
แล้วทั้งหมดนั้นจะหารด้วย
ดีที่ผมลอกและวางมันแล้ว
x บวก 1 คูณ x ลบ 1
ตอนนี้ นี่คือหัวใจของการกระจายเศษส่วนย่อย
เราบอกว่า โอเค เราทำมาทั้งหมดนี้
เพื่อเราจะได้อันนี้
ว่ามีค่า A กับ B ที่ทำให้อันนี้เป็นจริง
ถ้ามี A กับ B ที่อันนี้เป็นจริง
แล้ว A บวก B ต้องเป็นสัมประสิทธิ์
ของเทอม x นี่ตรงนี้
A บวก B ต้องเท่ากับ 1
ต้องเท่ากับสัมประสิทธิ์นี้
แล้ว B ลบ A ต้องเท่ากับค่าคงที่
ต้องเท่ากับลบ 4

Czech: 
Tohle přepíšeme jako…
Když vezmeme Ax plus Bx,
tak to bude (A plus B) krát x.
Pak máme -A a B.
Takže plus ,B' minus ,A'
a dám tam závorky,
aby to bylo hezky pohromadě.
A to všechno bude děleno…
Ještě že jsem si to zkopíroval.
(x plus 1) krát (x minus 1).
Tohle je podstata rozkladu
na parciální zlomky.
Dobře, prošli jsme si tímhle celým,
abychom našli nějaká
,A' a ,B', pro která to platí.
Takže jestli to platí
pro nějaká ,A' a ,B',
pak (A plus B) musí být
koeficient u členu s ,x'.
Takže (A plus B) musí být rovno 1,
musí to být rovno tomuto koeficientu.
No a (B minus A) musí být
rovno konstantě, tedy -4.

Portuguese: 
Podemos reescrever isso como,
se tomarmos Ax mais Bx
isto será A mais B vezes x.
Então temos um A negativo e um B.
Mais B menos A, e eu colocarei 
parênteses em torno disto
só para agrupar estes termos constantes.
Então, tudo isto será dividido por,
ainda bem que copiei e colei,
x mais um vezes x menos um.
Agora este é o ponto crucial da
expansão em frações parciais.
Dizemos ok, passamos por
todo o exercício na tese
que poderíamos fazer isso,
que existe algum A e
B para os quais isto é verdade.
Portanto, se há algum A e
B para os quais isto é verdade,
então A mais B deve ser o coeficiente
do termo x aqui.
Então, A mais B deve ser igual a um,
deve ser igual a este coeficiente.
E B menos A deve ser
igual à constante,
deve ser igual a quatro negativo.

English: 
that's going to be A plus B times X.
Then we have a negative A and a B.
So plus B minus A, and I'll
just put parenthesis around that
just so I kind of group
these constant terms.
Then all of that's going to be divided by,
good thing I copied and pasted that,
X plus one times X minus one.
So now this is the crux of
partial fraction expansion.
We say, okay we kind of went through this
whole exercise on the thesis
that we could do this,
that there is some A and
B for which this is true.
So if there is some A and
B for which this is true,
then A plus B must be the coefficient
of the X term right over here.
So A plus B must be equal to one,
must be equal to this coefficient.
And B minus A must be
equal to the constant,
must be equal to negative four.

Portuguese: 
Ou se eles são, então
acharemos um A e um B,
vamos fazer isso.
Farei isto aqui em cima já
que tenho pouco espaço.
A mais B será igual a um,
e B menos A ou eu poderia escrever
como A negativo mais B é
igual a quatro negativo.
Poderíamos adicionar o lado esquerdo
e o lado direito,
e então os A's desapareceriam.
Obteríamos dois B é
igual a três negativo
ou B é igual a 3/2 negativo.
Sabemos que A é igual a um menos B,
que seria igual a
um mais 3/2,
já que B é 3/2 negativo.
que é igual a 5/2.
A é igual a 5/2.
B é igual a 3/2 negativo.
E assim podemos
reescrever toda esta
integral de uma forma que
é um pouco mais fácil
de tomar a antiderivada,
ou toda esta expressão
portanto é mais fácil integrar.

Thai: 
ถ้าใช่ เราก็เจอ A กับ B
ลองทำกันดู
ผมจะทำบนนี้เพราะผมมีที่อยู่
A บวก B จะเท่ากับ 1
และ B ลบ A ผมเขียนมัน
ได้ลบ A บวก B เท่ากับลบ 4
เราบวกทางซ้ายมือ
และบวกทางขวามือได้
แล้ว A จะหายไป
เราจะได้ 2B เท่ากับลบ 3
หรือ B เท่ากับลบ 3/2
เรารู้ว่า A เท่ากับ 1 ลบ B
ซึ่งเท่ากับ 1 บวก 3/2
เพราะ B คือลบ 3/2
ได้เป็น 5/2
A เท่ากับ 5/2
B เท่ากับลบ 3/2
อย่างนั้น เราเขียนอินทิกรัลทั้งหมดนี้
ในรูปที่หาปฏิยานุพันธ์
ทั้งหมดนี้ได้
มันหาอินทิกรัลได้ง่ายขึ้น

English: 
Or if they are then we
will found an A and a B,
so let's do that.
I'll do it up here since I have
a little bit of real estate.
A plus B is going to be equal to one,
and B minus A or I could write that
as negative A plus B is
equal to negative four.
We could add the left hand sides
and add the right hand sides
and then the A's would disappear.
We would get two B is
equal to negative three
or B is equal to negative three halves.
We know that A is equal to one minus B,
which would be equal to
one plus three halves,
since B is negative three halves,
which is equal to five halves.
A is equal to five halves.
B is equal to negative three halves.
And just like that we
can rewrite this whole
integral in a way that
is a little bit easier
to take the anti or this whole expression
so it's easier to integrate.

Czech: 
Takže jestli to tak je, tak najdeme
ta ,A' a ,B', tak to pojďme udělat.
Udělám to tady, protože tu mám místo.
,A' plus ,B' bude rovno 1
a ,B' minus ,A', můžu taky
napsat -A plus B, bude rovno -4.
Můžeme sečíst levé strany a
pravé strany a pak ta ,A' zmizí.
Dostaneme, že 2B je rovno -3
neboli B je rovno -3/2.
Víme, že ,A' je rovno 1 minus B,
což je 1 plus 3/2, protože B je -3/2,
a to je rovno 5/2.
,A' je rovno 5/2.
,B' je rovno -3/2.
A takhle můžeme přepsat ten celý integrál,

Bulgarian: 
И ако това е така, 
ще намерим А и В,
хайде да го направим.
Ще го направя тук, защото
имам повече място.
(А + В) е равно на 1,
а (В – А)... ще използвам
различни цветове,
(–А + В) е равно на –4.
Можем да съберем
левите страни и десните страни,
и тогава А се унищожава с –А.
Остава 2В е равно на –3...
В е равно на –3/2.
Но ние знаем, че
А е равно на 1 – В,
което е равно на 1 + 3/2,
понеже В е –3/2,
или А е равно на 5/2.
В е равно на –3/2.
И сега можем да преработим
целия интеграл така, че
да стане малко по-лесно
да намерим примитивната
функция на този целия израз,
т.е. по-лесно да интегрираме.

Portuguese: 
Será a integral de
um mais A sobre x mais um.
A é 5/2 então posso
escrever isto como,
deixe-me escrever isto assim,
5/2 vezes um sobre x mais um.
Escrevi desta maneira porque
é mais simples
de tomar a antiderivada disto.
Em seguida, mais B sobre x menos um.
Que será 3/2 negativo.
Então vou escrever isto
como menos 3/2
vezes um sobre x menos um.
Foi isto aqui, dx.
Observe tudo o que fiz
foi tomar esta expressão
bem aqui e fiz um pouco
de expansão de fração parcial
nestas duas, acho você
poderia dizer, expressões
ou termos ali.

English: 
So it's going to be the integral of
one plus A over X plus one.
A is five halves and so I
could just write that as,
let me write it this way,
five halves times one over X plus one.
I wrote it that way because
it's very straight forward
to take the antiderivative of this.
Then plus B over X minus one.
Which is going to be
negative three halves.
So I'll just write it
as minus three halves
times one over X minus one.
That was this right over here, DX.
Notice all I did is I took this expression
right over here and I did a little bit
of partial fraction expansion
into these two, I guess
you could say, expressions
or terms right over there.

Thai: 
มันจะเท่ากับอินทิกรัลของ
1 บวก A ส่วน x บวก 1
A คือ 5/2 ผมก็เขียนมันได้เป็น
ขอผมเขียนแบบนี้นะ
5/2 คูณ 1 ส่วน x บวก 1
ผมเขียนแบบนั้นเพราะการหาปฏิยานุพันธ์
ของตัวนี้มันง่ายมาก
แล้วบวก B ส่วน x ลบ 1
ซึ่งจะเท่ากับลบ 3/2
ผมจะเขียนมันเป็นลบ 3/2
คูณ 1 ส่วน x ลบ 1
มันก็คือตรงนั้น dx
สังเกตว่าที่ผมทำคือผมนำพจน์นี้มา
ตรงนี้ แล้วผมทำการ
กระจายเศษส่วนย่อย
เป็นสองตัวนี้ คุณจะเรียกว่าพจน์
หรือเทอมก็ได้

Bulgarian: 
Получаваме интеграл от
1 + А/(х + 1).
А е равно на 5/2, така че
това ще стане...
ще го напиша по следния
начин.
5/2 по 1/(х + 1).
Написах го по този начин,
защото е много лесно
да намерим примитивната
функция на това.
После плюс В/(х –1).
В е равно на –3/2.
Ще го напиша като –3/2
по 1/(х – 1).
Това е ето това тук, dx.
Обърни внимание, че взех
този израз ето тук и просто
разложих елементарната дроб
на тези два, ако мога да 
кажа така, израза,
или члена ето тук.

Czech: 
aby bylo jednodušší to spočítat.
Takže to bude integrál z
1 plus (A děleno (x plus 1)).
,A' je 5/2, takže to napíšu
jako, napíšu to jako…
5/2 krát (1 děleno (x plus 1)).
Napsal jsem to tak, protože je
pak lehčí najít primitivní funkci.
Potom plus (B děleno (x minus 1)).
Což bude -3/2.
Takže to napíšu jako minus
(3/2 krát (1 děleno (x minus 1))).
To bylo tohle, a k tomu "dx".
Všimněte si, že jsem jen vzal tenhle výraz
a udělal jsem rozklad na
tyhle dva parciální zlomky,
na tyto výrazy nebo
členy, dalo by se říct.

Portuguese: 
É bastante simples
integrar isto.
Antiderivada de um,
será x.
Antiderivada de 5/2,
um sobre x mais um,
será mais 5/2,
o log natural do valor
absoluto de x mais um.
conseguimos fazer isto
pois a derivada
de x mais um é um,
então a derivada
está lá para que possamos
tomar a antiderivada
em relação a x mais um.
Você também pode fazer
substituição U como
nos exemplos anteriores,
U é igual a x mais um.
E aqui, isto será menos 3/2
vezes o log natural
do valor absoluto
de x menos um, pela
mesma lógica que
fomos capazes de tomar a
antiderivative lá.
E é claro, não podemos
esquecer nossa constante.
E aqui a temos.
Fomos capazes de integrar, conseguimos
avaliar essa expressão.
Legendado por [Raul Guimaraes]
[Revisado por Pilar Dib]

Czech: 
Je pak lehké integrovat tohle.
Primitivní funkce k 1 bude prostě x.
Primitivní funkce k 5/2
krát (1 děleno (x plus 1)),
to bude plus 5/2 krát přirozený logaritmus
z absolutní hodnoty z (x plus 1).
To můžeme udělat, protože
derivace z (x plus 1) je prostě 1,
takže ta derivace tu je
a můžeme to takhle udělat.
Mohli byste udělat i substituci
jako v předchozích videích,
,u' se rovná x plus 1.
A toto bude -3/2 krát přirozený logaritmus
z absolutní hodnoty z (x minus 1),
podle stejné logiky jako předtím.
A samozřejmě nemůžeme
zapomenout na konstantu.
A tady to máme.
Byli jsme schopní integrovat,
vyhodnotit tento výraz.

Bulgarian: 
Сега това е много лесно
за интегриране.
Примитивната функция на 1
е просто х.
Примитивната функция на 
5/2 по 1/(х + 1)
е равна на +5/2
по натурален логаритъм
от абсолютната стойност на (х + 1).
Можем да направим това,
защото производната
на (х + 1) е просто 1,
така че производната
е ето тук, така че можем
да намерим примитивната функция
спрямо (х + 1).
Можем да интегрираме със
заместване
като в предишния пример,
u = (х + 1).
И ето тук ще стане
–3/2
по натурален логаритъм
от абсолютната стойност
на (х – 1), по същата логика както
намерихме тази примитивна
функция ето тук.
И, разбира се, да не забравяме
нашата константа.
И това е.
Можахме да интегрираме,
успяхме да решим този израз.

Thai: 
มันอินทิเกรตได้ค่อนข้างง่าย
ปฏิยานุพันธ์ของ 1 มันก็แค่ x
ปฏิยานุพันธ์ของ 5/2, 1 ส่วน x บวก 1
จะเท่ากับบวก 5/2
ล็อกธรรมชาติของค่าสัมบูรณ์ของ x บวก 1
เราทำได้เพราะปฏิยานุพันธ์
ของ x บวก 1 ก็แค่ 1 อนุพันธ์
อยู่ตรงนั้น เราก็หาปฏิยานุพันธ์
เทียบกับ x บวก 1 ได้
คุณทำการแทนที่ u อย่างที่เราได้ทำ
ในตัวอย่างก่อนได้, u เท่ากับ x บวก 1
และตรงนี้ นี่คือลบ 3/2
คูณล็อกธรรมชาติของค่าสัมบูรณ์
ของ x ลบ 1, ด้วยเหตุผลเดียวกัน
เราหาปฏิยานุพันธ์ได้
และแน่นอน อย่าลืมค่าคงที่ของเรา
แล้วเราก็ได้คำตอบแล้ว
เราสามารถอินทิเกรต เราสามารถ
หาค่าพจน์นี้ได้แล้ว

English: 
It's fairly straight
forward to integrate this.
Antiderivative of one,
it's just going to be X.
Antiderivative of five
halves, one over X plus one,
is going to be plus five halves,
the natural log of the
absolute value of X plus one.
We're able to do that
because the derivative
of X plus one is just
one, so the derivative
is there so that we can
take the antiderivative
with respect to X plus one.
You could also do
u-substitution like we've done
in previous examples, U
is equal to X plus one.
And over here, this is going
to be minus three halves
times the natural log
of the absolute value
of X minus one, by the
same exact logic with how
we were able to take the
antiderivative there.
And of course we cannot
forget our constant.
And there we have it.
We've been able to integrate, we were able
to evaluate this expression.
