
English: 
Let's take the
indefinite integral
of the square root
of 7x plus 9 dx.
So my first question
to you is, is this
going to be a good case
for u-substitution?
Well, when you look here,
maybe the natural thing
to set to be equal
to u is 7x plus 9.
But do I see its derivative
anywhere over here?
Well, let's see.
If we set u to be
equal to 7x plus 9,
what is the derivative of u
with respect to x going to be?
Derivative of u
with respect to x
is just going to be equal to 7.
Derivative of 7x is 7.
Derivative of 9 is 0.
So do we see a 7 lying
around anywhere over here?
Well, we don't.
But what could we do in order
to have a 7 lying around,
but not change the
value of the integral?
Well, the neat thing-- and we've
seen this multiple times-- is
when you're
evaluating integrals,

Portuguese: 
Vamos calcular a integral indefinida
da raiz quadrada de sete vezes x mais nove dx.
Minha primeira pergunta é: esse é um bom 
caso para uma substituição de u?
Olhando aqui, talvez o mais natural seja
fazer u igual a sete vezes x mais nove,
mas vemos sua derivada 
em algum lugar por aqui?
Vejamos, se fizermos u
igual a sete vezes x mais nove
qual é a derivada de u em relação a x?
A derivada de u em relação a x
será igual a sete
A derivada de sete vezes x é sete, 
a derivada de nove é zero.
Estamos vendo algum sete por aqui?
Não. Mas o que poderíamos fazer
para ter um sete por aqui
mas sem mudar o valor da integral?
Uma coisa interessante que vimos muitas 
vezes é que quando lidamos com integrais,

Thai: 
ลองหาอินทิกรัลไม่จำกัดเขตของสแควรืรูทของ 7x บวก 9 dx กัน
คำถามแรกให้คุณคือว่า มันเหมาะกับการแทนตัว u หรือเปล่า?
เวลาคุณดูตรงนี้, เราก็ทำตามธรรมชาติ ให้ u เท่ากับ 7x บวก 9,
แต่ผมเห็นอนุพันธ์ของมันอยู่ตรงไหนในนี้หรือเปล่า?
ลองดู, เราตั้ง u เท่ากับ 7x บวก 9, แล้วอนุพันธ์ของ u เทียบกับ x จะเป็นเท่าไหร่?
อนุพันธ์ของ u เทียบกับ x จะเท่ากับ 7
อนุพันธ์ของ 7x เป็น 7, อนุพันธืของ 9 ได้ 0. เราเห็น 7 อยู่ในนี้ตรงไหนไหม?
ที่จริง, เราไม่เห็น. แต่เราทำอะไรได้เพื่อให้มี 7 อยู่ด้วย, โดยไม่เปลี่ยนค่าของอินทิกรัล?
ทีนี้, สิ่งที่เจ๋งและเราเห็นมาแล้วครั้งแล้วเวลาคุณหาค่าอินทิกรัล,

Polish: 
Policzmy całkę nieoznaczoną z pierwiastka kwadratowego z 7x+9 dx.
Moje pierwsze pytanie brzmi: czy to dobry przykład na całkowanie przez podstawienie?
Jeśli tu popatrzycie, możecie pomyśleć, że naturalnym pomysłem będzie oznaczenie jako u 7x+9
ale czy widać gdzieś tu pochodną tego wyrażenia?
Zobaczmy, oznaczamy u jako 7x+9, jaka będzie pochodna u względem x?
Pochodna u względem x to po prostu będzie 7.
Pochodna 7x to 7, pochodna 9 to 0. Czy zatem widzimy gdzieś tu siódemkę?
Cóż, niezbyt. Ale co moglibyśmy zrobić, żeby siódemka się pojawiła, ale wartość całki się nie zmieniła?
Cóż, fajną rzeczą - i widywaną wiele razy, kiedy liczymy całki

Korean: 
 
√(7x+9)의 부정적분을
계산해 봅시다
제 첫 질문은 이 문제가 과연
u-치환에 적절한 예시일까요?
문제를 보시면, u를 7x+9로
설정하는 것이 자연스러운 것 같습니다
그런데 이 문제의 어디에서 도함수를 찾죠?
글쎄요 한번 봅시다
만약 u를 7x+9로 설정하면
x에 관한 도함수는 무엇이 될까요?
x에 관한 도함수는
바로 7이 될 것입니다
7x의 도함수는 7이고
9의 도함수는 0이니까요
그럼 이 문제에서 7을 찾을 수 있나요?
아뇨 그러지 못하네요
과연 정적분의 값을 안바꾸면서도
7을 찾기 위해 어떻게 해야 할까요?
우리가 많이 봐왔으며 가장 중요한 것은
정적분을 계산할 때

Bulgarian: 
Нека намерим неопределения
интеграл
от квадратен корен от 7 по х
плюс 9, dx.
Първият ми въпрос към теб е:
дали тук е подходящо да
интегрираме със заместване?
Когато разглеждаме израза, може би
естественото нещо
е да положим (заместим)
u да е равно на 7 по х плюс 9.
Виждаме ли обаче производната
му някъде тук?
Нека да видим.
Ако положим u да е равно
на 7 по х плюс 9,
на какво ще бъде равна
производната на u спрямо х?
Производната на u спрямо х
ще бъде равна на 7.
Производната на 7 по х
е равна на 7.
Производната на 9 е равна на 0.
Виждаме ли 7 да се намира
някъде тук?
Не, не виждаме.
Какво може да направим обаче,
за да получим 7 под интеграла,
но без да променяме стойността
на интеграла?
Има едно хубаво нещо, което сме
виждали множество пъти.
При изчисляване на интеграли,
скаларните величини (числа)
могат да бъдат

Czech: 
Neurčitý integrál
druhé odmocniny z (7x plus 9) dx.
Má první otázka zní,
lze použít substituci?
Přirozeně by nás napadlo
označit 7x plus 9 jako 'u',
ale vidíme tu derivaci tohoto výrazu?
Podívejme se, bude-li 'u' rovno 7x plus 9,
jaká bude derivace 'u' podle 'x'?
Derivace 'u' podle 'x' by byla 7.
Derivace 7x je 7, derivace 9 je 0.
Vidíme tu tedy někde 7?
Nevidíme.
Ale co můžeme udělat,
abychom ji tu měli
a zároveň nezměnili hodnotu integrálu?

Turkish: 
7 x artı 9'un karekökünün integralini alıyoruz.
Size ilk sorum şu. Burada yerine koyma yöntemini uygulayabilir miyiz?
Bu soruya baktığımızda, u eşittir 7 x artı 9 demek istiyoruz.
Ancak, bu ifadenin türevini herhangi bir yerde görüyor muyum?
Eğer u eşittir 7 x artı 9 dersek, u'nun x'e göre türevi ne olur?
u'nun x'e göre türevi 7'dir.
7 x'in türevi 7, 9'un türevi de 0. Peki, burada 7 görüyor muyuz?
Görmüyoruz. İntegralin değerini değiştirmeden buraya nasıl 7 koyabiliriz?
İntegral alırken gördüğümüz çok güzel bir sonuç skalerleri integralin dışına alabilmemizdi.

Turkish: 
-
Hatırlamak istersek, a skaleri çarpı f x d x'in integrali eşittir a çarpı integral f x d x.
-
Yani skaler ile fonksiyonun çarpımının integrali eşittir skaler çarpı fonksiyonun integrali. Bunu şuraya yazayım.
-
Buna göre, içinde 7 olan bir ifadeyle integrali çarpıp bölebilir miyiz?
Evet, 7 ile çarpıp bölebiliriz. Orijinal integrali tekrardan yazayım.
-
Orijinal integrali, 1 bölü 7 çarpı 7 çarpı 7 x artı 9'un karekökü d x'in integrali olarak yazabiliriz.
-

Polish: 
liczby mogą bardzo łatwo wchodzić lub wychodzić spod całki.
Przypominam, że jeśli mamy całkę z
powiedzmy jakiaś liczba a razy jakaś funkcja f(x) dx, to jest to samo co
a razy całka z f(x)dx. Całka z liczby razy funkcja jest równe
liczba razy całka z funkcji. Zostawię to tu z boku.
Wiedząc to, czy możemy przez coś pomnożyć i podzielić, żeby pokazało się 7?
Cóż, możemy pomnożyć i podzielić przez 7. Wyobraźcie to sobie teraz. Przepiszmy naszą wyjściową całkę
narysuję strzałkę, żeby nie wchodzić na to, co już napisałem.
Możemy przepisać naszą wyjściową całkę jako równą całce
z 1/7 razy 7 razy pierwiastek z 7x+9 dx

Portuguese: 
multiplicadores podem facilmente
entrar e sair da integral.
Vamos lembrar que se temos a integral
um fator a vezes f de dx, é a mesma coisa
que a vezes a integral de f de x dx.
A integral de um fator vezes uma função
é igual ao escalar
vezes a integral da função.
Com isso em mente, podemos multiplicar e 
dividir por algo de forma a obter um sete?
Podemos multiplicar e dividir por sete. 
Vamos reescrever nossa integral original
Podemos reescrever nossa integral 
original como sendo
a integral de 1/7 vezes sete vezes a raiz
quadrada de 7 vezes x mais nove dx.

Korean: 
스칼라는 적분기호 안과 밖을 
쉽게 오갈 수 있다는 것입니다
만약 a 스칼라 배의 f(x)dx의
정적분을 계산할 때
이는 f(x)dx의 정적분 값의 a배와 같습니다
함수의 스칼라 배의 정적분과
함수의 정적분 값을 스칼라 배 한 것은 같습니다
이것을 바로 옆에 두겠습니다
이를 사용해, 우리가 무언가로 곱하고 나누어
7이 나타나게 할 수 있을까요?
글쎄요, 우리는 7을 곱하고 나눌 수 있습니다
이것을 상상해보세요
원래 정적분 문제를 다시 쓰겠습니다
밑으로 넘어가기 위해
작은 화살표를 하나 그리겠습니다
우리는 원래 정적분에 1/7과 7을 곱하고
뒤에 √(7x+9) dx를 붙여서
다시 쓸 수 있습니다

Thai: 
สเกลาร์สามารถเข้าออกในอินทิกรัลได้สบาย
เพื่อเตือนตัวเองว่าถ้าผมมีอินทิกรัลของ,
สมมุติว่าสเกลาร์ a คูณ f(x) (f ของ x) dx, นี่ก็เหมือนกับ
a คูณอินทิกรัลของ f(x) dx. นั่นคืออินทิกรัลของสเกลาร์คูณฟังก์ชัน เท่ากับ
สเกลาร์นั้นคูณอินทิกรัลของฟังก์ชัน. แล้วขอผมใส่นี่ไว้ข้างๆ ตรงนี้นะ
เมื่อรู้แล้ว, เราคูณและหารด้วยอะไรถึงจะมี 7 โผล่ขึ้นมา?
ทีนี้, เราคูณและหารด้วย 7. แล้วจินตนาการดูตรงนี้. ลองเขียนอินทิกรัลอันเดิมใหม่ --
ขอผมวาดลูกศรเล็กๆ ตรงนี้ไปข้างๆ นี้
เราสามารถเขียนอินทิกรัลเดิมของเรา ว่าเท่ากับอินทิกรัล
ของ 1/7 คูณ 7 คูณสแควร์รูทของ 7x บวก 9 dx

English: 
scalars can go in and outside
of the integral very easily.
Just to remind ourselves, if I
have the integral of let's say
some scalar a times
f of x dx, this
is the same thing as a times
the integral of f of x dx.
The integral of the
scalar times a function
is equal to the scalar times
the integral of the functions.
So let me put this
aside right over here.
So with that in mind, can
we multiply and divide
by something that will
have a 7 showing up?
Well, we can multiply
and divide by 7.
So imagine doing this.
Let's rewrite our
original integral.
So let me draw a
little arrow here just
to go around that aside.
We could rewrite our
original integral
as being 9 to the integral
of times 1/7 times 7 times
the square root of 7x plus 9 dx.

Bulgarian: 
внасяни или изнасяни от интеграла.
Нека само да си припомним.
Ако имам интеграл
от някакво число а, умножено
по f от x, dx,
това е равно на същото като а,
умножено по интеграл от f от x, dx.
Интеграл от число, умножено
по функция,
е равно на числото, умножено
по интеграл от функцията.
Нека да отделя това ето така.
Като вземем това предвид,
можем ли да умножим и разделим
на нещо този интеграл, така че под
интеграла да се появи 7?
Може да умножим
и разделим на 7.
Нека си представим следното.
Нека запишем по друг начин
първоначалния интеграл.
Ще нарисувам една стрелка,
за да пиша направо отдолу.
Може да запишем
първоначалния интеграл
като равен на интеграл от 1/7,
умножено по 7,
умножено по квадратен корен
от 7 по х плюс 9, dx.
Може да изнесем числото 1/7

Czech: 
U integrálů je dobré to, že čísla
se snadno dostávají z a do integrálů.
Pro připomenutí,
máme-li integrál 'a' krát f(x) dx,
je to stejné jako
'a' krát integrál f(x) dx.
Násobím-li integrand nějakým číslem,
mohu to číslo „vytknout“ před integrál.
Dám to tedy bokem.
Víme-li to, můžeme to něčím vydělit
a vynásobit tak, abychom tu dostali 7?
Můžeme to vynásobit a vydělit 7.
Přepišme tedy původní integrál.
Nakreslím tu šipku, abychom to tu obešli.
Integrál z (1 lomeno 7) krát 7
krát odmocnina z (7x plus 9) dx.

Korean: 
1/7을 정적분의 밖으로
뺄 수 있습니다
그럴 필요는 없지만 다시 쓰자면
이것은 7√(7x+9) dx의 정적분 값의
1/7배가 됩니다
u를 7x+9로 설정했으면
이제 우리는 도함수를 찾을 수 있나요?
당연하죠
7은 바로 여기 있거든요
du를 다른 방식으로 쓰면
du는 7dx입니다
그래서 du는 dx의 7배이죠
이 부분들은 du와 같습니다
u에 관해서 신경을 쓰고 싶다면
그것은 바로 7x+9입니다
이것이 u입니다
그럼 u에 관한 문자로 이 부정적분을 다시 써보죠
이것은 정적분의 1/7배와 같은데요
7을 맨 뒷부분에 놓겠습니다
그러면 √u du라고 다시 쓸 수 있으며

English: 
And if we want to, we
could take the 1/7 outside
of the integral.
We don't have to,
but we can rewrite
this as 1/7 times
the integral of 7,
times the square
root of 7x plus 9 dx.
So now if we set u
equal to 7x plus 9,
do we have its
derivative laying around?
Well, sure.
The 7 is right over here.
We know that du-- if we want to
write it in differential form--
du is equal to 7 times dx.
So du is equal to 7 times dx.
That part right over
there is equal to du.
And if we want to care
about u, well, that's
just going to be the 7x plus 9.
That is are u.
So let's rewrite this indefinite
integral in terms of u.
It's going to be equal to
1/7 times the integral of--
and I'll just take the 7
and put it in the back.
So we could just
write the square root

Turkish: 
İstersek 1 bölü 7'yi dışarı alırız.
Böyle yapmak zorunda değiliz, ama bunu 1 bölü 7 çarpı, 7 çarpı karekök 7 x artı 9 d x'in integrali olarak yazabiliriz.
-
Şimdi u eşittir 7 x artı 9 dersek, bunun türevini görüyor muyuz?
Tabii ki! Diferansiyel şeklinde yazmak istersek d u eşittir 7 çarpı d x deriz. d u eşittir 7 çarpı d x.
Buradaki kısım d u'ya eşit.
u da 7 x artı 9 olacak.
Bu, u olarak belirttiğimiz ifade. Şimdi tüm integrali u cinsinden yazalım.
1 bölü 7 çarpı - buradaki 7'yi arkaya yazacağım, böylece karekök u d u'nun integrali diyebiliriz. 7 çarpı d x eşittir d u.
-

Polish: 
I jeśli chcemy, możemy wyciągnąć 1/7 przed całkę
nie musimy, ale możemy to przepisać jako
1/7 razy całka z 7 razy pierwiastek kwadratowy z 7x+9 dx.
Zatem, gdybyśmy powiedzieli "u jest równe 7x+9", czy widać gdzieś tu jego pochodną?
Pewnie! 7 jest tutaj. Wiemy, że du, jeśli chcemy to zapisać w formie różniczkowej
du jest równe 7 razy dx. Czyli du jest równe 7 dx. Ta część tutaj jest równa du
i widzimy gdzie jest nasze u, będzie to po prostu 7x+9.
To jest nasze u. Przepiszmy więc naszą całkę jako wyrażenie od u.
Będzie ona równa 1/7 razy całka z
i przerzucę 7 na koniec, żebyśmy mogli po prostu napisać

Portuguese: 
Se quisermos, podemos tirar 1/7
para fora da integral.
Não temos que, mas podemos reescrever
1/7 vezes a integral de sete vezes a raiz
quadrada de sete vezes x mais nove dx.
Agora, se u é igual a sete vezes x 
mais nove, temos sua derivada?
Sim! O sete está bem aqui. Sabemos que se 
quisermos escrever du na forma diferencial
du é igual a sete vezes dx. 
Esta parte aqui é igual a du
e quando pensamos em u, 
será igual a sete vezes x mais nove.
Esse é nosso u. Vamos reescrever
essa integral indefinida em termos de u.
Será igual a 1/7 vezes a integral
Vou trazer o 7 para trás, assim
podemos escrever

Czech: 
(1 lomeno 7) tedy můžeme
vytknout před integrál, chceme-li.
(1 lomeno 7) krát integrál z
7 krát odmocnina (7x plus 9) dx.
Pokud nyní bude 'u' rovno 7x plus 9,
máme tu někde derivaci tohoto výrazu?
Jistě! 7 máme přímo zde,
víme, že 'du' ve tvaru diferenciálu…
'du' je rovno 7 krát dx.
Tato část je rovna 'du'.
'u' bude 7x plus 9.
Přepišme integrál do proměnných 'u'.
Bude to (1 lomeno 7) krát integrál z…

Thai: 
และถ้าเราต้องการ, เราสามารถนำ 1/7 ออกจากอินทิกรัลได้,
เราไม่ต้องก็ได้, แต่เราสามารถเขียนนี่เป็น
1/7 คูณอินทิกรัลของ 7 คูณสแควร์รูทของ 7x บวก 9 dx
ทีนี้, ถ้าเราบอกว่า, "u เท่ากับ 7x บวก 9" แล้วเราเห็นอนุพันธ์ของมันอยู่ตรงไหนไหม?
แน่นอน! 7 อยู่ตรงนี้. เรารู้ว่า du, ถ้าเราเขียนมันในรูปดิฟเฟอเรนเชียล,
du เท่ากับ 7 คูณ dx. du จึงเท่ากับ 7 คูณ dx. ส่วนนั่นตรงนั้นเท่ากับ du,
และถ้าเราอยากได้ u, มันจะเป็น 7x บวก 9
นั่นคือ u ของเรา. แล้วลองเขียนอินทิกรัลทั้งหมดนี่ในรูปของ u ดู
มันจะเท่ากับ 1/7 คูณอินทิกรัลของ --
ผมก็นำ 7 มาไว้ข้างหลัง แล้วเราก็สามารถเขียน

Bulgarian: 
извън интеграла.
Не е задължително, но сега
може да го запишем като
1/7, умножено по интеграл от 7
по квадратен корен от
7 по х плюс 9, dx.
Сега може да положим (заместим)
7 по х плюс 9 да е равно на u.
Имаме ли производната на този
израз под интеграла?
Разбира се!
Ето тази седмица тук.
Знаем, че du – ако искаме да го
запишем с диференциали –
е равно на 7 по dx.
И така, du е равно на 7 по dx.
Тази част ето тук
е равна на du.
Ако искаме да покажем
къде се намира u,
то това просто ще бъде
изразът 7 по х плюс 9.
Това е нашето u.
Нека запишем този неопределен
интеграл, изразен чрез u.
Ще бъде равно на 1/7,
умножено по интеграл...
ще запиша седмицата последна...

Portuguese: 
a raiz quadrada de u du.
Sete vezes dx é igual a du.
Se quisermos podemos reescrever como u 
elevado a 1/2. Isso facilita um pouco
para aplicarmos a regra da exponencial. 
Então podemos reescrever
1/7 vezes a integral de u elevado a 1/2 du
Esse u pode ficar em branco, da mesma cor
que esse du aqui.
Qual é a anti-derivada de u elevado a 1/2?
Aumentamos a potência de u em um, 
então isso será igual a
--não podemos nos esquecer 
desse 1/7 aqui fora--
Então será igual a 1/7 vezes --se
incrementamos a potência aqui, será

Thai: 
สแควร์รูทของ u du. 7 คูณ dx เท่ากับ du
แล้ว, เราสามารถเขียนนี่ใหม่ได้ ถ้าเราอยากได้เป็น u ยกกำลัง 1/2, ทำให้มัน
ง่ายขึ้นหน่อยเวลาใช้กฎยกกำลังย้อนกลับตรงนี้. เราก็สามารถเขียนนี่ใหม่ว่าเท่ากับ
1/7 คูณอินทิกรัลของ u ยกกำลัง 1/2 du. แล้วขอผมบอกให้ชัดหน่อย
u นี่ผมเขียนมันด้วยสีขาว, ผมอยากได้สีเดียวกับ du นี่ตรงนี้ มันเหมือนกัน
แล้วแอนติเดริเวทีฟของ u กำลัง 1/2 เป็นเท่าไหร่?
ทีนี้, เราเพิ่มกำลัง u ขึ้น 1, มันจะเท่ากับ --
ผมยังไม่ลืม 1/7 นี้ข้างหน้านะ
มันจะเท่ากับ 1/7 คูณ -- ถ้าเราเพิ่มกำลังตรงนี้, มันจะเป็น

Polish: 
pierwiastek z u du. 7 razy dx to du.
I, jeśli chcemy, możemy to przepisać jako u do potęgi 1/2, łatwiej nam wtedy zobaczyć
że tak naprawdę całkujemy funkcję potęgową, co umiemy robić. Przepisujemy to zatem
1/7 razy całka z u do potęgi 1/2 du. I postaram się teraz wyjaśnić.
To u mogłem napisać na biało. Chcę ten sam kolor du, bo jest to to samo du, co tutaj.
Więc, jaka jest całka z u do potęgi 1/2?
Cóż, zwiększamy wykładnik o 1, więc będzie to równe
i nie mogę zapomnieć o 1/7 na początku.
Czyli będzie to 1/7 razy, jeśli zwiększymy wykładnik o 1, będzie to równe

English: 
of u du, 7 times dx is du.
And we can rewrite this if we
want as u to the 1/2 power.
It makes it a little bit
easier for us to kind of do
the reverse power rule here.
So we can rewrite this as equal
to 1/7 times the integral of u
to the 1/2 power du.
And let me just make it clear.
This u I could have
written in white
if I want it the same color.
And this du is the same
du right over here.
So what is the antiderivative
of u to the 1/2 power?
Well, we increment
u's power by 1.
So this is going to be
equal to-- let me not
forget this 1/7 out front.
So it's going to be 1/7 times--
if we increment the power here,
it's going to be u to the 3/2,
1/2 plus 1 is 1 and 1/2 or 3/2.
So it's going to
be u to the 3/2.

Turkish: 
-
Bunu u üzeri 1 bölü 2 olarak yazabiliriz. Böylece kuvvet kuralının tersini daha kolay uygularım.
-
Bunu 1 bölü 7 çarpı u üzeri 1 bölü 2 d u'nun integrali olarak yazabiliriz.
-
u üzeri 1 bölü 2'nin terstürevi nedir?
u'nun üssünü 1 artırırız.
Öndeki 1 bölü 7'yi unutmayalım.
Yani 1 bölü 7 çarpı u üzeri 3 bölü 2 (1 bölü 2 artı 1 eşittir 1 tam 1 bölü 2 veya 3 bölü 2) ve bu yeni ifadeyi 3 bölü 2'nin çarpmaya göre tersiyle çarpıyoruz, ki bu da 2 bölü 3'tür.

Bulgarian: 
по интеграл от квадратен корен
от u, du, което е 7 по x, dx.
Може да запишем този интеграл
като u на степен 1/2.
Така ще е по-лесно да
използваме
правилото за намиране
на примитивната функция.
Ще бъде равно на 1/7,
умножено по интеграл от u
на степен 1/2, du.
Нека да го обясня.
Това u ще го запиша с бяло.
Нека да са еднакъв цвят.
Това du също, защото е равно
на това du тук.
Коя е примитивната функция
на u на степен 1/2?
Увеличаваме степента
на u с единица.
Тогава този интеграл е равен
на следното.
Нека не забравяме, че има 1/7
изнесено отпред.
Получава се 1/7, умножено... ако
увелича степента с 1
ще се получи u на степен 3/2, което
е равно на 1 плюс 1/2...
по u на степен 3/2.

Czech: 
7 dám na konec, 7 dx je 'du'.
Píšu odmocninu z 'u' krát 'du'.
Odmocninu z 'u' mohu přepsat na
'u na (1 lomeno 2)', je to pak snazší.
Bude to rovno (1 lomeno 7) krát integrál z
'u na (1 lomeno 2)' krát 'du'.
Označím to barevně stejně jako předtím.
Jaká je primitivní funkce k
'u na (1 lomeno 2)'?
Zvyšujeme mocninu 'u' o 1,
bude to tedy…
Nesmím zapomenout na (1 lomeno 7).

Korean: 
dx의 7배는 du입니다
이것을 다시 1/2제곱을 바꿔 쓸 수 있습니다
이것은 우리가 적분 규칙을 쓰기
쉽게 만들어 줍니다
그래서 우리는 이 식을 u 1/2제곱 du의
정적분 값의 1/7배라고 쓸 수 있습니다
더 깔끔하게 만들어보죠
같은 색으로 쓰기 위해서
이 u를 하얀색을 쓸게요
du는 이 du와 같습니다
√u의 부정적분은 뭘까요?
우리는 u의 지수를 1 증가시켜서
1/7을 앞으로 빼놓는 것을
잊지 않으면서
1/7배가 될 것입니다
여기서 지수를 증가시키면
1/2더하기 1인 3/2가 u의 지수가 됩니다
그래서 u의 3/2이 됩니다

Polish: 
u do potęgi 3/2 (1/2 plus 1 to półtora, czyli 3/2) i musimy pomnożyć to nowe wyrażenie przez odwrotność 3/2
czyli 2/3. I zachęcam was, byście sprawdzili, że pochodna 2/3 razy u do 3/2 to istotnie u do potęgi 1/2.
Mamy już to i skoro mnożymy to wszystko przez 1/7
możemy tu jeszcze dorzucić plus C, które może być pewną stałą
i jeśli chcemy, możemy pomnożyć nawias przez 1/7. Dostaniemy wtedy 1/7 razy 2/3 to 2/21 razy u^(3/2)
i 1/7 razy stała, co da nam po prostu jakąś inną stałą
mogę zatem napisać tu stałą. Tamtą mogę nazwać C1, zaś tą mogę nazwać C2
ale tak naprawdę to po prostu jakaś stała. I gotowe, och, właściwie to jeszcze nie.
Mamy odpowiedź, ale jako wyrażenie od u. Musimy jeszcze odwrócić podstawienie.

Turkish: 
-
2 bölü 3 çarpı u üzeri 3 bölü 2'nin türevinin gerçekten de u üzeri 1 bölü 2 olduğunu kontrol etmenizi isteyeceğim.
Bunu 1 bölü 7 ile çarpıyoruz ve artı C sabitini şuraya ekliyoruz.
-
İstersek 1 bölü 7'yi dağıtabiliriz. Yani 1 bölü 7 çarpı 2 bölü 3 eşittir 2 bölü 21 u üzeri 3 bölü 2.
Ve 1 bölü 7 çarpı bir sabit, bu da başka bir sabittir.
Buna C 1, şuna C 2 diyebilirim.
-
Cevabımız u cinsinden olduğu için u yerine x'li ifadeyi koymamız gerekiyor.

Portuguese: 
u elevado a 3/2 (1/2 mais 1 
é um e 1/2 ou 3/2).
Vamos multiplicar esse termo pelo
recíproco de 3/2, que é 2/3
Recomendo verificar que a derivada de 2/3
vezes u elevado a 3/2 é u elevado a 1/2.
Como estamos multiplicando 1/7 vezes
essa integral indefinida
podemos colocar mais c aqui, que é
uma constante.
Vamos distribuir esse 1/7. Então temos 
1/7 vezes 2/3 que é 2/21 u elevado a 3/2
e 1/7 vezes uma constante, que será 
apenas uma constante.
Podemos chamar uma de C1 e esta C2
mas são apenas constantes arbitrárias.
Ainda temos nossa equação em termos de u.
Então temos que desfazer a substituição.

English: 
And then we're going to
multiply this new thing
times the reciprocal
of 3/2, which is 2/3.
And I encourage you to verify
the derivative of 2/3 u
to the 3/2 is
indeed u to the 1/2.
And so we have that.
And since we're
multiplying 1/7 times
this entire indefinite
integral, we
could also throw in a
plus c right over here.
There might have
been a constant.
And if we want, we can
distribute the 1/7.
So it would get 1/7 times
2/3 is 2/21 u to the 3/2.
And 1/7 times some
constant, well, that's
just going to be some constant.
And so I could write
a constant like that.
I could call that c1 and
then I could call this c2,
but it's really just
some arbitrary constant.
And we're done.
Oh, actually, no we aren't done.
We still just have our
entire thing in terms of u.
So now let's unsubstitute it.

Czech: 
(1 lomeno 7) krát 'u na (3 lomeno 2)',
to vynásobíme převrácenou hodnotou,
tedy krát (2 lomeno 3).
Chci abyste si ověřili, že derivace
(2 lomeno 3) krát 'u na (3 lomeno 2)
je opravdu 'u na (1 lomeno 2)'.
Jelikož násobíme celý integrál,
přidáme sem i konstantu C.
Roznásobme (1 lomeno 7).
(1 lomeno 7) krát (2 lomeno 3)
je (2 lomeno 21),
(1 lomeno 7) krát konstanta
je nějaká jiná konstanta.
Mohl bych je od sebe rozlišit indexem,
ale je to prostě jen nějaká konstanta.
A jsme hotovi…
Vlastně nejsme.
Stále máme výsledek v proměnné 'u',
dosaďme tedy za 'u'.

Bulgarian: 
u на степен 3/2.
Сега следва да умножим
този нов израз
по реципрочната стойност на 3/2,
която е 2/3.
Насърчавам те да провериш,
че производната
на 2/3 по u на степен 3/2 наистина
е равна на u на степен 1/2.
Това е, което се получава.
Тъй като умножаваме по 1/7
целия този интеграл,
може да прибавим и една
константа C ето тук.
Възможно е тук
да има константа.
Може и да разкрием скобите и да
умножим с числото 1/7.
Получава се 1/7 по 2/3 – което
е равно на 2/21 – по u на степен 3/2.
А 1/7, умножено по произволна
константа,
отново ще бъде някаква
произволна константа.
Мога да я запиша и ето така.
Тази да означа с C1, а тази с C2,
но наистина това е просто
някаква константа.
И сме готови!
О, всъщност не сме готови.
Все още имаме целия този израз
като функция на u.
Нека да заместим
положеното за u.

Korean: 
 
그리고 이 수에 3/2의 역수인
2/3을 곱합니다
그리고 (2/3)u의 3/2제곱의 도함수가
u의 1/2 제곱임을 확인해보세요
이렇게 값을 얻었습니다
전체 부정적분에 1/7배를 할 때부터
우리는 괄호 안으로
c를 집어 넣을 수 있습니다
상수가 있어야 하니까요
1/7을 분배할 수도 있습니다
결과적으로 (2/21)u의 3/2제곱을 얻습니다
(1/7)c는 그냥 다른 상수로
바뀌어도 됩니다
이런식으로 그 상수를 써보겠습니다
전 상수를 c1, 이 상수를 c2라고 하죠
그냥 임의의 상수일 뿐입니다
그럼 끝났네요
오, 아직 안끝났네요
아직 u에 관한 식이 남아있네요
치환을 해제해 봅시다

Thai: 
u กำลัง 3/2 (1/2 บวก 1 ได้ 1 1/2 หรือ 3/2), และเราจะคูณเจ้าตัวใหม่นี่ด้วยส่วนกลับของ 3/2,
ซึ่งก็คือ 2/3, และผมแนะนำให้คุณทดสอบว่า อนุพันธ์ของ 2/3 * u^(3/2) เท่ากับ u^(1/2) จริง
แล้วเราก็ได้นั่นมา และเนื่องจากเราคูณ 1/7 เข้ากับอินทิกรัลไม่จำกัดเขตทั้งหมดนี่
เราก็โยนบวก c เข้าไปตรงนี้เป็นค่าคงที่,
และถ้าเราต้องการ, เราสามารถกระจาย 1/7 ได้. แล้วมันจะได้ 1/7 คูณ 2/3 ได้ (2/21) u^(3/2),
คูณ 1/7 คูณค่าคงที่, ทีนี้มันจะเป็นค่าคงที่สักค่า,
แล้วเราก็เขียนค่าคงที่แบบนั้น, ผมจะเรียกมันว่า c1, และผมเรียกนี่ว่า c2 ได้,
แต่มันเป็นค่าคงที่อะไรก็ได้. แล้วเราก็เสร็จแล้ว -- โอ้, ที่จริงเรายังไม่เสร็จ
เรายังต้องมีทุกอย่างในรูปของ u. ตอนนี้ลองแทนค่าย้อนกลับกัน

Thai: 
นี่ก็จะเท่ากับ 2/21 คูณ u^(3/2) -- และเรารู้แล้วว่า u เท่ากับอะไร,
u เท่ากับ 7x บวก 9. ขอผมใช้สีใหม่หน่อยเพราะมันน่าเบื่อแล้ว
มันจะเท่ากับ 2/21 คูณ 7x บวก 9 ยกกำลัง 3/2 บวก C
แล้วเราก็เสร็จแล้ว! เราสามารถจัดการเจ้าอินทิกรัลยุ่งเหยิงนี่ได้ แล้ว
สังเกตว่า แม้มันจะไม่ชัดเจนนัก แต่การแทนตัว u ใช้ได้ในกรณีนี้

Bulgarian: 
Целият този израз е равен на
2/21 по u на степен 3/2.
Знаем на какво е равно u.
u е равно на 7х плюс 9.
Нека да използвам друг цвят,
за да се отличава.
Получава се 2/21, умножено по
7х плюс 9,
на степен 3/2, плюс C.
И вече сме готови!
Успяхме да решим този страховито
изглеждащ интеграл,
и въпреки че не беше напълно
очевидно в началото
все пак да открием и приложим
интегриране със заместване.

Czech: 
(2 lomeno 21) krát 'u na (3 lomeno 2)'
a víme, že 'u' je (7x plus 9).
Použiji novou barvu,
aby to nebylo tak jednotvárné.
(2 lomeno 21) krát
'(7x plus 9) na (3 lomeno 2)' plus C.
A jsme hotovi!
Tento integrál jsme, ačkoliv to nebylo
ihned zřejmé, vyřešili pomocí substituce.

Portuguese: 
Isso é igual a 2/21 vezes u elevado a 3/2.
Já sabemos o valor de u
u é igual a sete vezes x mais nove.
Então temos 2/21 vezes sete vezes x
mais nove elevado a 3/2 mais c.
Terminamos essa integral cabeluda
e vimos que mesmo não sendo óbvio
conseguimos aplicar a substituição por u.
[Legendado por Eduardo Roder]
[Revisado por: Pilar Dib]

Polish: 
Będzie to zatem 2/21 razy u do (3/2), a wiemy, czemu jest równe u,
u jest równe 7x+9. Użyję nowego koloru, żeby złagodzić trochę monotonię.
Czyli będzie to równe 2/21 razy 7x+9 do potęgi 3/2 plus C.
I gotowe! Jesteśmy w stanie wziąć dość brzydko wyglądającą całkę i zdać sobie sprawę
iż, mimo że to nie było od początku tak oczywiste, całkowanie przez podstawienie tutaj działa.

Turkish: 
2 bölü 21 çarpı u üzeri 3 bölü 2 ve u'nun neye eşit olduğunu biliyoruz. u eşittir 7 x artı 9.
-
Yani cevabimiz 2 bölü 21 çarpı 7 x artı 9'un 3 bölü 2'nci kuvveti artı C'dir.
Ve cevabı bulduk. Karmaşık görünen bir integralde yerine koyma yöntemini kullanmayı başardık.
-

Korean: 
(2/21)u의 3/2제곱을 구할 때
우리는 u가 무엇인지 이미 압니다
u는 바로 7x+9이죠
단조로움을 없애기 위해 새로운 색을 넣겠습니다
그래서 이 수는
(2/21)*(7x+9)의 (3/2)제곱 +c가 됩니다
 
그럼 됐네요
처음에 u-치환이 적용되는 것 같지 않아
명백해보이지 않더라도
어려운 적분을 풀 수 있게 되었습니다
 

English: 
So this is going to be equal
to 2/21 times u to the 3/2.
And we already know
what u is equal to.
u is equal to 7x plus 9.
Let me put a new color here
just to ease the monotony.
So it's going to be
2/21 times 7x plus 9
to the 3/2 power plus c.
And we are done.
We were able to take a kind
of hairy looking integral
and realize that even
though it wasn't completely
obvious at first, that
u-substitution is applicable.
