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Let's take the indefinite integral of the square root of 7x plus 9 dx.

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So my first question to you is is this going to be a good case for u-substitution?

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When you look here, we maybe the natural thing to set to be equal to u is 7x plus 9,

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but do I see its derivative anywhere over here?

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Let's see, we set u to be equal to 7x plus 9, what is the derivative of u with respect to x going to be?

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Derivative of u with respect to x is just going to be equal to 7.

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Derivative of 7x is 7, derivative of 9 is 0. So do we see a seven lying around anywhere over here?

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Well, we don't. But what would could we do in order to have a seven lying around, but not change the value of the integral?

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Well, the neat thing and we see this multiple times is when you're evaluating integrals,

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

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Just to remind ourselves that if I have the integral of,

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let's say some scalar a times f(x) (f of x) dx, this is the same thing as

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a times the integral of f(x) (f of x) dx. That the integral of the scalar times the function is equal to

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the scalar times the integral of the function. So let me put this aside right over here.

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So with that in mind, can we multiply and divide by something that will have a 7 showing up?

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Well, we can multiply and divide by 7. So imagine doing this. Let's rewrite our original integral --

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so let me draw a little arrow here just to go around that aside.

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We could write our original integral as being equal to the integral

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of 1/7 times 7 times the square root of 7x plus 9 dx.

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And if we want to, we could take the 1/7 outside of the integral,

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we don't have to, but we could rewrite this as

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1/7 times the integral of 7 times the square root of 7x plus 9 dx.

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So now, if we said, "u is equal to 7x plus 9," do we have its derivative laying around?

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Well, sure! The 7 is right over here. We know that du, if we want to write it in differential form,

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du is equal to 7 times dx. So du is equal 7 times dx. That part right there is equal to du,

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and if we want to care about u, well that's just going to be the 7x plus 9.

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That is our u. So let's rewrite this entire integral in terms of u.

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

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and I'll just take the 7 and put it in the back so we could just write

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the square root of u du. Seven times dx is du.

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And, we can rewrite this if we want as u to the 1/2 power; makes it a little bit easier

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for us to kind of do the reverse power rule here. So we can rewrite this is equal to

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1/7 times the integral of u to the 1/2 power du. And let me just make it clear.

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This u I could have written it in white; I want the same color in this du is the same du right over here.

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So what is the anti-derivative of u to the 1/2 power?

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Well, we increment u's power by 1, so this is going to be equal to --

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and let me not forget this 1/7 out front.

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

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u to the 3/2 (1/2 plus 1 is 1 1/2 or 3/2), and we're going to multiply this new thing times the reciprocal of 3/2,

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which is 2/3. And I encourage you to verify that derivative 2/3 * u^(3/2) is indeed u^(1/2).

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And so we have that and since we're multiplying 1/7 times this entire indefinite integral,

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we could also throw in a plus C right over here that might be a constant,

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and if we want, we can distribute the 1/7. So it would get 1/7 times 2/3 is (2/21)u^(3/2),

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and 1/7 times some constant, well that's just going to be some constant,

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so I could write a constant like that; I could call it C1, and I could call this C2,

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but it really is some arbitrary constant. And we're done -- oh, actually no we aren't done.

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We still just have our entire thing in terms of u. So now let's un-substitute it.

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So this is going to be equal to 2/21 times u^(3/2) -- and we already know what u is equal to,

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u is equal to 7x plus 9. Let me put in a new color here just to ease the monotony.

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So it's going to be 2/21 times 7x plus 9 to the 3/2 power plus C.

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And we are done! We're able to take a kind of hairy looking integral and realize

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that even though it wasn't completely obvious at first, u-substitution is applicable.