Question

Evaluate: anti derivative

Square root (3x+9x) dx

would the answer be square root (3/2x^2+9x) + c

Answer 1

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

Answer 2

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

Answer 3

Is that what you have?

Answer 4

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

Answer 5

Ah. Have you done u substitutions yet?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

u = 3x+9 :)

Answer 12

..+C...

Answer 13

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

Answer 14

amistre64 you've the patient of Job!

Answer 15

lol....thanx ;)

Answer 16

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.

Answer 17

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

Answer 18

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.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

\[f(g(x)) = f(3x+9) = \sqrt{3x+9}\]

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

You're missing a factor of 2 on the front. And your exponent is flipped the wrong way.

It should be:
\[\int \frac{\sqrt{u}}{3}du \]
\[= \int \frac{u^{1/2}}{3} du\]
\[= \frac{2}{3}*\frac{u^{3/2}}{3} + C\]
\[= \frac{2(3x+9)^{3/2}}{9} + C\]

Answer 28

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???

Answer 29

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.

Answer 30

Where a and C are constants.

Answer 31

I am going to start a new question

Answer 32

Ok!