I wanted to take the derivative of the anti-derivative just to see if it checks out, but I'm not really sure how to simplify what I got; I got (x-1)^6 + 4(x-1)^5. What's the best way to simplify that so it turns back into the original function in the integral?

Factor an ( x-1 )^5 out and you get ( ( x-1 )+4 )( x-1 )^5 which is equal to ( x+3 )( x-1 )^5, the original function.

integral of x√x-1dx please help me !

Maybe you could substitute u=(x-1), then du=dx and x=u+1. You get ∫[u+1]√u du. Distributive property en linearity of integrals yields the following: ∫u√u du + ∫√u du equals ∫u^3/2 du + ∫u^(½) du… Which comes down to integrating basic polynomials.

How did Sal know to substitute u+1 for x?

when you do u-subs, you want to turn whatever is the most complicated part of the problem (in this case (x-1)^5) into a simpler form so it will be easier. The general 'rule' for doing this is to make u equal to whatever is inside whatever is making it complex (in this case, x-1 is inside, and the ^5 is what makes it complex), so u=x-1. Also, whenever u is equal to ax + b, there will never be any variables created when you do the u-sub, so you don't have to worry about getting rid of those variables.

What if they both had exponents? For example: integral (x+3)^2 times (x-1)^3. How would you solve this? Thanks

it's possible to substitute x-1 = u, then x = u+1, we put both in. Then you can expand (u+4)^2 and it's pretty simple from there.

what is the solution for x/root of x+4

∫(𝑥∕√(𝑥 + 4))𝑑𝑥 Let 𝑢 = √(𝑥 + 4) ⇒ 𝑥 = 𝑢² − 4 ⇒ ⇒ 𝑥∕√(𝑥 + 4) = (𝑢² − 4)∕𝑢 = 𝑢 − 4∕𝑢 = = √(𝑥 + 4) − 4∕√(𝑥 + 4) Thereby, ∫(𝑥∕√(𝑥 + 4))𝑑𝑥 = ∫√(𝑥 + 4)𝑑𝑥 − ∫(4∕√(𝑥 + 4))𝑑𝑥 Let 𝑣 = 𝑥 + 4 ⇒ 𝑑𝑣 = 𝑑𝑥 Thereby, ∫(𝑥∕√(𝑥 + 4))𝑑𝑥 = ∫√𝑣 ∙ 𝑑𝑣 − 8 ∙ ∫(1∕(2√𝑣))𝑑𝑣 = = (2∕3)√𝑣³ − 8√𝑣 = (2∕3)√(𝑥 + 4)³ − 8√(𝑥 + 4)

I understand the video in general, but why is it called "back substituting"?

I'm really not sure why he called it back substituting. Saying that `x+3=u+1+3=u+4` just seems like a normal old substitution for me. I would think that the back substitution would be the last step in which you get your antiderivative in terms of x. Just remember: if you do u substitution, you need all of your x's in the integrand to disappear.

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𝘶-substitution: special application

Name: 𝘶-substitution: special application
Uploaded: 2012-12-28T22:32:58Z
Description: Using 𝘶-substitution in a situation that is a bit different than "classic" 𝘶-substitution. In this case, the substitution helps us take a hairy expression and make it easier to expand and integrate.

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Using 𝘶-substitution in a situation that is a bit different than "classic" 𝘶-substitution. In this case, the substitution helps us take a hairy expression and make it easier to expand and integrate. Created by Sal Khan.

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Zach Mertens
Posted 11 years ago. Direct link to Zach Mertens's post “I wanted to take the deri...”
I wanted to take the derivative of the anti-derivative just to see if it checks out, but I'm not really sure how to simplify what I got; I got (x-1)^6 + 4(x-1)^5. What's the best way to simplify that so it turns back into the original function in the integral?
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(33 votes)
Answer
- Saul
  Posted 11 years ago. Direct link to Saul's post “Factor an ( x-1 )^5 out a...”
  Factor an ( x-1 )^5 out and you get ( ( x-1 )+4 )( x-1 )^5 which is equal to ( x+3 )( x-1 )^5, the original function.
  Button navigates to signup page
  (48 votes)
Jakob.delaval
Posted 11 years ago. Direct link to Jakob.delaval's post “Why is it more difficult ...”
Why is it more difficult doing integral by parts in this case? Maybe I've missed something essential, but isn't the integral of 5(x-1)^4 =(x-1)^5 ? Well, I guess it's not, since I get a wrong answer.
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(4 votes)
Answer
- little ghost
  Posted 4 months ago. Direct link to little ghost's post “because the (x-1)^5 isn’t...”
  because the (x-1)^5 isn’t alone, it’s being multiplied by the (x+3), so it’s difficult to take the antiderivative as it is
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  (1 vote)
Flavour
Posted 11 years ago. Direct link to Flavour's post “I did this, it's kinda th...”
I did this, it's kinda the same just without substitutions which makes it a bit more clear that you can re-write expressions very easy for many problems. (x+3)=(x+3-4+4)=(x-1+4)
so (x-1+4)(x-1)^5 just multiply to get (x-1)^6+4(x-1)^5 now it's easy ∫(x-1)^6+4(x-5)^5 dx=1/7(x-1)^7+2/3(x-1)^6+C
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(6 votes)
Answer
Janice
Posted 11 years ago. Direct link to Janice 's post “How did Sal know to subst...”
How did Sal know to substitute u+1 for x?
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(3 votes)
Answer
- Derek Edrich
  Posted 11 years ago. Direct link to Derek Edrich's post “when you do u-subs, you w...”
  when you do u-subs, you want to turn whatever is the most complicated part of the problem (in this case (x-1)^5) into a simpler form so it will be easier. The general 'rule' for doing this is to make u equal to whatever is inside whatever is making it complex (in this case, x-1 is inside, and the ^5 is what makes it complex), so u=x-1. Also, whenever u is equal to ax + b, there will never be any variables created when you do the u-sub, so you don't have to worry about getting rid of those variables.
  Button navigates to signup page
  (5 votes)
Donald Hajderasi
Posted 10 years ago. Direct link to Donald Hajderasi's post “integral of x√x-1dx pleas...”
integral of x√x-1dx please help me !
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(4 votes)
Answer
- Zeta
  Posted 10 years ago. Direct link to Zeta's post “Maybe you could substitut...”
  Maybe you could substitute u=(x-1), then du=dx and x=u+1. You get ∫[u+1]√u du. Distributive property en linearity of integrals yields the following: ∫u√u du + ∫√u du equals ∫u^3/2 du + ∫u^(½) du… Which comes down to integrating basic polynomials.
  Comment on Zeta's post “Maybe you could substitut...”
  (2 votes)
jbanful
Posted 11 years ago. Direct link to jbanful's post “What if they both had exp...”
What if they both had exponents? For example: integral (x+3)^2 times (x-1)^3. How would you solve this? Thanks
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(3 votes)
Answer
- Mantas Vaskevicius
  Posted 11 years ago. Direct link to Mantas Vaskevicius's post “it's possible to substitu...”
  it's possible to substitute x-1 = u, then x = u+1, we put both in. Then you can expand (u+4)^2 and it's pretty simple from there.
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  (2 votes)
Murtaza Salihi
Posted 8 years ago. Direct link to Murtaza Salihi's post “Why not just use integrat...”
Why not just use integration by part here?
Button navigates to signup pageComment on Murtaza Salihi's post “Why not just use integrat...”
(2 votes)
Answer
Peter Kapeel
Posted 11 years ago. Direct link to Peter Kapeel's post “when one is given an inte...”
when one is given an integral problem, in what order should I try to use the different techniques that we have learned.

i.e. should I always try u sub first, integration by parts second or what method order would you suggest?
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(2 votes)
Answer
ASHISH TIWARI
Posted 5 years ago. Direct link to ASHISH TIWARI's post “what is the solution for ...”
what is the solution for x/root of x+4
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(1 vote)
Answer
- Jerry Nilsson
  Posted 5 years ago. Direct link to Jerry Nilsson's post “∫(𝑥∕√(𝑥 + 4))𝑑𝑥 Let ...”
  ∫(𝑥∕√(𝑥 + 4))𝑑𝑥
  
  Let 𝑢 = √(𝑥 + 4) ⇒ 𝑥 = 𝑢² − 4 ⇒
  ⇒ 𝑥∕√(𝑥 + 4) = (𝑢² − 4)∕𝑢 = 𝑢 − 4∕𝑢 =
  = √(𝑥 + 4) − 4∕√(𝑥 + 4)
  
  Thereby,
  ∫(𝑥∕√(𝑥 + 4))𝑑𝑥 = ∫√(𝑥 + 4)𝑑𝑥 − ∫(4∕√(𝑥 + 4))𝑑𝑥
  
  Let 𝑣 = 𝑥 + 4 ⇒ 𝑑𝑣 = 𝑑𝑥
  
  Thereby,
  ∫(𝑥∕√(𝑥 + 4))𝑑𝑥 = ∫√𝑣 ∙ 𝑑𝑣 − 8 ∙ ∫(1∕(2√𝑣))𝑑𝑣 =
  = (2∕3)√𝑣³ − 8√𝑣 = (2∕3)√(𝑥 + 4)³ − 8√(𝑥 + 4)
  Button navigates to signup page
  (3 votes)
jyuan
Posted 10 years ago. Direct link to jyuan's post “I understand the video in...”
I understand the video in general, but why is it called "back substituting"?
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(1 vote)
Answer
- Ethan Dlugie
  Posted 10 years ago. Direct link to Ethan Dlugie's post “I'm really not sure why h...”
  I'm really not sure why he called it back substituting. Saying that x+3=u+1+3=u+4 just seems like a normal old substitution for me. I would think that the back substitution would be the last step in which you get your antiderivative in terms of x.
  Just remember: if you do u substitution, you need all of your x's in the integrand to disappear.
  Button navigates to signup page
  (2 votes)

Video transcript

We're faced with the indefinite integral of x plus 3, times x minus 1 to the fifth dx. Now, we could solve this by literally multiplying out what x minus 1 to the fifth is. Maybe using the binomial theorem, that would take a while. And then we'd multiply that times x plus 3, and we'd end up with some polynomial. And we could take the antiderivative that way. Or we could maybe make a substitution here that could simplify this expression here. Make it something that's a little bit easier to take the antiderivative of. And this isn't going to be kind of the more traditional u substitution, where we just set u equal to something and see if its derivative is there. But it's a kind of a form of u substitution, where we do set u equal to something and see if it simplifies our expression in a way that, I guess, simplifies it. So let's try things out. So we have this x minus 1 to the fifth. That would be a pain to expand out. It would be nice if this was just a u to the fifth. So let's just set this, let's just set this to be equal to u. So let's set u as equal to x minus 1. And in that case, du is equal to dx. We could write du/dx is equal to 1, derivative of x, derivative of negative 1 is just 0. And these two are completely equivalent statements. And so if we did that, how could we rewrite this entire expression? Well, it would be equal to the integral of-- well, we have x plus 3 right over here, this is neither just u nor is it du. So let's think about how, what we can do here. Well, we could say, if u is equal to x minus 1, we could add 1 to both sides of this equation. And we could say u plus 1 is equal to x. And so for x, we can substitute that with u plus 1. So let's do that. So we're kind of back substituting in for x. So x is equal to u plus 1. And I'm just trying to see if I can do something to simplify this expression. So x is u plus 1, then we have our plus 3 there, times x minus 1 to the fifth. x minus 1 was u, that's the simplification we wanted to make. So times u to the fifth power. And dx is the same thing as du. So du. Now did this get us anywhere? Can we simplify this to a form that it's easy to take the antiderivative of? Well, it looks like it did. Let's see. We can rewrite this as this expression right over here is just u plus 4. Times u to the fifth-- I'll do it all in one color, now-- du. And the reason why this simplified things-- the way this simplified things is taking x minus one to the fifth, that'd be a really hard thing to expand. But u to the fifth is just u to the fifth. And then it just changed this x plus 3 into a u plus 4, which is still a pretty straightforward expression. And now we can just distribute the u to the fifth. So we are left with u to the sixth power plus 4, u to the fifth du. And this is a pretty straightforward thing to take the antiderivative of. Now you might be saying, hey Sal, this was-- how did you know to set u equal to be that. And oftentimes with integration, it's going to take a little bit of trial and error. There's a certain bit of an art to it. But here the realization was well, x minus 1 to the fifth can be really complicated. Maybe u to the fifth might make it a little bit simpler. And that did just happen to work. You could have tried u is equal to x plus 3. But it wouldn't have simplified it as nicely as u equal to x minus 1 did. But let's finish with this integral right over here. So this is going to be equal to the antiderivative of u to the sixth. Well, that's just u to the seventh over seven. Plus the antiderivative of u to the fifth, that's u to the sixth over six. But we have the 4 out here, so it's 4 times u to the sixth over six. And then we have a plus C. And this, the 4u 6 is the same thing as 2/3, so we can rewrite this whole thing as equal to u to the seventh over seven, plus 2/3 u to the sixth, plus C. And now we just have to undo our u substitution. u is equal to x minus 1. So this is going to be equal to x minus 1 to the seventh over 7 plus 2/3 times x minus 1 to the sixth, plus C. And we're all done. We were able to take a fairly hairy-- or what could have been a hairy thing if we had to expand this out-- and we were able to take the antiderivative by doing a little bit of this u substitution and u back substitution.

Calculus 2

Course: Calculus 2 > Unit 2

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