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Let's take the[br]indefinite integral

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of the square root[br]of 7x plus 9 dx.

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So my first question[br]to you is, is this

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going to be a good case[br]for u-substitution?

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Well, when you look here,[br]maybe the natural thing

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to set to be equal[br]to u is 7x plus 9.

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But do I see its derivative[br]anywhere over here?

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Well, let's see.

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If we set u to be[br]equal to 7x plus 9,

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what is the derivative of u[br]with respect to x going to be?

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Derivative of u[br]with respect to x

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is just going to be equal to 7.

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Derivative of 7x is 7.

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Derivative of 9 is 0.

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So do we see a 7 lying[br]around anywhere over here?

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Well, we don't.

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But what could we do in order[br]to have a 7 lying around,

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but not change the[br]value of the integral?

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Well, the neat thing-- and we've[br]seen this multiple times-- is

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when you're[br]evaluating integrals,

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scalars can go in and outside[br]of the integral very easily.

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Just to remind ourselves, if I[br]have the integral of let's say

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some scalar a times[br]f of x dx, this

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is the same thing as a times[br]the integral of f of x dx.

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The integral of the[br]scalar times a function

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is equal to the scalar times[br]the integral of the functions.

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So let me put this[br]aside right over here.

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So with that in mind, can[br]we multiply and divide

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by something that will[br]have a 7 showing up?

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Well, we can multiply[br]and divide by 7.

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So imagine doing this.

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Let's rewrite our[br]original integral.

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So let me draw a[br]little arrow here just

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to go around that aside.

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We could rewrite our[br]original integral

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as being 9 to the integral[br]of times 1/7 times 7 times

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the square root of 7x plus 9 dx.

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And if we want to, we[br]could take the 1/7 outside

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of the integral.

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We don't have to,[br]but we can rewrite

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this as 1/7 times[br]the integral of 7,

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times the square[br]root of 7x plus 9 dx.

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So now if we set u[br]equal to 7x plus 9,

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do we have its[br]derivative laying around?

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Well, sure.

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The 7 is right over here.

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We know that du-- if we want to[br]write it in differential form--

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du is equal to 7 times dx.

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So du is equal to 7 times dx.

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That part right over[br]there is equal to du.

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And if we want to care[br]about u, well, that's

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just going to be the 7x plus 9.

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That is are u.

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So let's rewrite this indefinite[br]integral in terms of u.

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It's going to be equal to[br]1/7 times the integral of--

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and I'll just take the 7[br]and put it in the back.

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So we could just[br]write the square root

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of u du, 7 times dx is du.

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And we can rewrite this if we[br]want as u to the 1/2 power.

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It makes it a little bit[br]easier for us to kind of do

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the reverse power rule here.

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So we can rewrite this as equal[br]to 1/7 times the integral of u

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to the 1/2 power du.

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And let me just make it clear.

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This u I could have[br]written in white

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if I want it the same color.

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And this du is the same[br]du right over here.

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So what is the antiderivative[br]of u to the 1/2 power?

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Well, we increment[br]u's power by 1.

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So this is going to be[br]equal to-- let me not

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forget this 1/7 out front.

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So it's going to be 1/7 times--[br]if we increment the power here,

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it's going to be u to the 3/2,[br]1/2 plus 1 is 1 and 1/2 or 3/2.

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So it's going to[br]be u to the 3/2.

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And then we're going to[br]multiply this new thing

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times the reciprocal[br]of 3/2, which is 2/3.

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And I encourage you to verify[br]the derivative of 2/3 u

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to the 3/2 is[br]indeed u to the 1/2.

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And so we have that.

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And since we're[br]multiplying 1/7 times

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this entire indefinite[br]integral, we

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could also throw in a[br]plus c right over here.

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There might have[br]been a constant.

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And if we want, we can[br]distribute the 1/7.

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So it would get 1/7 times[br]2/3 is 2/21 u to the 3/2.

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And 1/7 times some[br]constant, well, that's

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just going to be some constant.

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And so I could write[br]a constant like that.

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I could call that c1 and[br]then I could call this c2,

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but it's really just[br]some arbitrary constant.

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And we're done.

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Oh, actually, no we aren't done.

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We still just have our[br]entire thing in terms of u.

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So now let's unsubstitute it.

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So this is going to be equal[br]to 2/21 times u to the 3/2.

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And we already know[br]what u is equal to.

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u is equal to 7x plus 9.

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Let me put a new color here[br]just to ease the monotony.

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So it's going to be[br]2/21 times 7x plus 9

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to the 3/2 power plus c.

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And we are done.

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We were able to take a kind[br]of hairy looking integral

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and realize that even[br]though it wasn't completely

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obvious at first, that[br]u-substitution is applicable.

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