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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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• 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.
• 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.
• 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
(1 vote)
• 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
• 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.
• 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.
• 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.
• Why not just use integration by part here?
• 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?
• I understand the video in general, but why is it called "back substituting"?
(1 vote)
• 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.
∫ ((x+3) (x-1)^5) dxLet (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))) dxby 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) = 1g'(x) = e^(5 ln(x-1)) g(x) = d/dx(e^(5 ln(x-1))) = [ e^(5 ln(x-1)) * (x - 1) ] / 5then ∫ ((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) ]/ 5computer's answer of the problem : x^7 / 7 - x^6 / 3 - x^5 + 5x^4 - 25x^3 / 3 + 7x^2 - 3x