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𝘶-substitution: special application

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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  • purple pi purple style avatar for user Zach Mertens
    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?
    (33 votes)
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  • blobby green style avatar for user Jakob.delaval
    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.
    (4 votes)
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  • leaf orange style avatar for user Flavour
    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
    (6 votes)
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  • leaf red style avatar for user Janice Coleman
    How did Sal know to substitute u+1 for x?
    (3 votes)
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    • leaf green style avatar for user Derek Edrich
      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.
      (5 votes)
  • blobby green style avatar for user Donald Hajderasi
    integral of x√x-1dx please help me !
    (4 votes)
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    • aqualine seed style avatar for user Zeta
      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.
      (2 votes)
  • leafers seedling style avatar for user jbanful
    What if they both had exponents? For example: integral (x+3)^2 times (x-1)^3. How would you solve this? Thanks
    (3 votes)
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  • leafers seed style avatar for user Murtaza Salihi
    Why not just use integration by part here?
    (1 vote)
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  • starky ultimate style avatar for user Peter Kapeel
    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?
    (2 votes)
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  • male robot hal style avatar for user jyuan
    I understand the video in general, but why is it called "back substituting"?
    (1 vote)
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    • spunky sam blue style avatar for user Ethan Dlugie
      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.
      (2 votes)
  • piceratops ultimate style avatar for user Mina
    Hey,
    I tried this problem on my own and saw that I could solve it by raising e to some power :
    ∫ ((x+3) (x-1)^5) dx

    Let (x-1) = e^(ln(x-1))

    then (x-1)^5 = e^(5 ln(x-1))

    then ∫ ((x+3) (x-1)^5) dx = ∫ ((x+3) e^(5 ln(x-1))) dx

    by applying the integration by parts :

    ┌─────────────────────────────────────────────┐
    │∫(f(x)•g'(x))dx = f(x)•g(x) - ∫(f'(x)•g(x))dx│
    └─────────────────────────────────────────────┘

    f(x) = (x + 3)

    f'(x) = d/dx(x + 3) = 1

    g'(x) = e^(5 ln(x-1))

    g(x) = d/dx(e^(5 ln(x-1))) = [ e^(5 ln(x-1)) * (x - 1) ] / 5

    then ∫ ((x+3) e^(5 ln(x-1))) dx = (x + 3) * [ e^(5 ln(x-1)) * (x - 1) ] / 5 - [ e^(5 ln(x-1)) * (x - 1) ] / 5

    = [ (x - 1)^6 * (x + 2) ]/ 5
    computer's answer of the problem : x^7 / 7 - x^6 / 3 - x^5 + 5x^4 - 25x^3 / 3 + 7x^2 - 3x

    When I graph this function on a grapher and Sal's answer and computer's answer of the problem I get some similarity between them
    Please someone help me out.
    (1 vote)
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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.