## anonymous 5 years ago Evaluate: anti derivative Square root (3x+9x) dx would the answer be square root (3/2x^2+9x) + c

1. amistre64

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

2. anonymous

$\sqrt{3x+9x}$

3. anonymous

Is that what you have?

4. anonymous

$\sqrt{3x + 9} dx$

5. anonymous

Ah. Have you done u substitutions yet?

6. amistre64

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

7. amistre64

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

8. amistre64

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

9. anonymous

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

10. amistre64

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

11. amistre64

u = 3x+9 :)

12. amistre64

..+C...

13. amistre64

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

amistre64 you've the patient of Job!

15. amistre64

lol....thanx ;)

16. 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.

17. anonymous

18. 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.

19. amistre64

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

20. amistre64

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

21. amistre64

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

22. 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 $f(g(x)) = f(3x+9) = \sqrt{3x+9}$

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. 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???

29. 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.

30. anonymous

Where a and C are constants.

31. anonymous

I am going to start a new question

32. anonymous

Ok!