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Let's take the
indefinite integral

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of the square root
of 7x plus 9 dx.

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So my first question
to you is, is this

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going to be a good case
for u-substitution?

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Well, when you look here,
maybe the natural thing

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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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Well, let's see.

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

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

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Derivative of u
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
around anywhere over here?

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Well, we don't.

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

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but not change the
value of the integral?

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

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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, if I
have the integral of let's say

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some scalar a times
f of x dx, this

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

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The integral of the
scalar times a function

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is equal to the scalar times
the integral of the functions.

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So let me put this
aside right over here.

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So with that in mind, can
we multiply and divide

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by something that will
have a 7 showing up?

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Well, we can multiply
and divide by 7.

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So imagine doing this.

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Let's rewrite our
original integral.

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So let me draw a
little arrow here just

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to go around that aside.

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We could rewrite our
original integral

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as being 9 to the integral
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
could take the 1/7 outside

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of the integral.

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We don't have to,
but we can rewrite

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this as 1/7 times
the integral of 7,

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times the square
root of 7x plus 9 dx.

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

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do we have its
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
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
there is equal to du.

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And if we want to care
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
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.

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So we could just
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
want as u to the 1/2 power.

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It makes it a little bit
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
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
written in white

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if I want it the same color.

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And this du is the same
du right over here.

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

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Well, we increment
u's power by 1.

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So this is going to be
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--
if we increment the power here,

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

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

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And then we're going to
multiply this new thing

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times the reciprocal
of 3/2, which is 2/3.

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

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to the 3/2 is
indeed u to the 1/2.

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And so we have that.

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And since we're
multiplying 1/7 times

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this entire indefinite
integral, we

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could also throw in a
plus c right over here.

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There might have
been a constant.

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

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

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

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just going to be some constant.

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And so I could write
a constant like that.

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I could call that c1 and
then I could call this c2,

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but it's really just
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
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
to 2/21 times u to the 3/2.

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And we already know
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
just to ease the monotony.

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So it's going to be
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
of hairy looking integral

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and realize that even
though it wasn't completely

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obvious at first, that
u-substitution is applicable.

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