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## Calculus 2

### Unit 2: Lesson 1

Integrating with u-substitution- 𝘶-substitution intro
- 𝘶-substitution: multiplying by a constant
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: defining 𝘶 (more examples)
- 𝘶-substitution
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: rational function
- 𝘶-substitution: logarithmic function
- 𝘶-substitution warmup
- 𝘶-substitution: indefinite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution with definite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution: definite integral of exponential function
- 𝘶-substitution: special application
- 𝘶-substitution: double substitution
- 𝘶-substitution: challenging application

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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.

## Want to join the conversation?

- 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)
- 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.(47 votes)

- 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)
- 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) - How did Sal know to substitute u+1 for x?(3 votes)
- 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)

- integral of x√x-1dx please help me !(4 votes)
- 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)

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

- Why not just use integration by part here?(1 vote)
- 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) - 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.(2 votes)

- 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)

## 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.