Evaluate: anti derivative
Square root (3x+9x) dx
would the answer be square root (3/2x^2+9x) + c

- anonymous

Evaluate: anti derivative
Square root (3x+9x) dx
would the answer be square root (3/2x^2+9x) + c

- katieb

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- amistre64

sqrt(12x) .... is that correct?

- anonymous

\[\sqrt{3x+9x}\]

- anonymous

Is that what you have?

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## More answers

- anonymous

\[\sqrt{3x + 9} dx\]

- anonymous

Ah. Have you done u substitutions yet?

- amistre64

(3x + 9)^1/2 you need a 3, so multiply it by 3/3 and draw out the (1/3)

- amistre64

(1/3) [S] 3 (3x+9)^1/2

- amistre64

1/3 [S] u^1/2 du is what it becomes

- anonymous

no I have not
I am having trouble with this section. Not sure when I need to substitute

- amistre64

2sqrt(u^3)/9 if I did it right

- amistre64

u = 3x+9 :)

- amistre64

..+C...

- amistre64

after that last "bout" my brain is acting like a neanderthal

- radar

amistre64 you've the patient of Job!

- amistre64

lol....thanx ;)

- anonymous

As a u substitution you would just say
\[u = 3x+9 \implies du = 3dx \implies dx = \frac{du}{3}\]
Then
\[\int \sqrt{3x+9}\ dx = \int \sqrt{u} * \frac{1}{3}du\]
Then take the anti derivative of the u version and when you are done substitute back in 3x+9 for u.

- anonymous

ok but why will I use substitution. What about this problem tells me to substitute?

- amistre64

that "3x+9" under the radical is the clue. in order to derive down to this state, we had to use the chain rule. And that part is now missing from this equation; we essentially need to get it back.

- amistre64

Dx(3x+9) = 3 ..... so we need to account for a "missing" 3

- amistre64

if u = 3x+9 ; then du = 3 dx and dx= du/3

- amistre64

sqrt(u) du
-------- is what we can work out easily
3

- anonymous

When you are finding anti-derivatives of a function composed of another function
\[\int f(g(x)) dx\] You will want to use a u substitution.
In this case \[f(a) = \sqrt{a}\]
and
\[g(x) = 3x+9\]
And you're finding the anti-derivative f(g(x) dx
\[f(g(x)) = f(3x+9) = \sqrt{3x+9}\]

- anonymous

is the answer (3x+9)^2/3 / 9 +C

- anonymous

Not quite. What did you get for the anti derivative of \[\frac{1}{3}\sqrt{u}\ du?\]

- anonymous

Actually I see where your mistake was. You need to bring down the reciprocal of the exponent _after_ you add 1 to it.

- anonymous

Actually, it looks like you did that. I think it was just a book keeping mistake.

- anonymous

You're missing a factor of 2 on the front. And your exponent is flipped the wrong way.
It should be:
\[\int \frac{\sqrt{u}}{3}du \]
\[= \int \frac{u^{1/2}}{3} du\]
\[= \frac{2}{3}*\frac{u^{3/2}}{3} + C\]
\[= \frac{2(3x+9)^{3/2}}{9} + C\]

- anonymous

I see what you are doing but it just isn't clicking for me, but I am working on it. The 2/3 * u^3/2/3 The 2/3 is just the recp of 3/2???

- anonymous

Yes. Recall that the anti-derivative of
\[x^a \implies \frac{1}{a+1}*x^{a+1} + C\]
We can confirm this is true, by taking the derivative of the right side and verifying that we get the left side.

- anonymous

Where a and C are constants.

- anonymous

I am going to start a new question

- anonymous

Ok!

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