1. ## U-Substitution question...

Q:

2. Originally Posted by qbkr21
Q:
you are not wrong!

Some algebraic manipulation should get you the right answer. if you need help with the simplification, say so

3. Originally Posted by qbkr21
Q:
$\displaystyle \frac {(1 + 2x)^{ \frac {3}{2}} - 3 \sqrt {1 + 2x}}{6}$

$\displaystyle = \frac { \sqrt {1 + 2x} [(1 + 2x) - 3]}{6}$

$\displaystyle = \frac { \sqrt {1 + 2x} (2x - 2)}{6}$

$\displaystyle = \frac { \sqrt {1 + 2x} (x - 1) \cdot 2}{6}$

$\displaystyle = \frac {(x - 1) \sqrt {1 + 2x}}{3}$

4. ## Re:

Hey thanks Jhevon! So any time you do try to solve for "x" and it works out you take that du value which in this case was $\displaystyle \frac{1}{2}$ and simplify it along with the rest of the equation? I guess what I am trying to ask is... you DO NOT MOVE THE DU OUT IN FRONT, that would be bad wouldn't it? Thanks again for all of the tutoring tonight...

Thanks!

-qbkr21

5. Originally Posted by qbkr21
Hey thanks Jhevon! So any time you do try to solve for "x" and it works out you take that du value which in this case was $\displaystyle \frac{1}{2}$ and simplify it along with the rest of the equation? I guess what I am trying to ask is... you DO NOT MOVE THE DU OUT IN FRONT, that would be bad wouldn't it? Thanks again for all of the tutoring tonight...

Thanks!

-qbkr21
no you don't move the du out in front, keep it at the end. i always move my constants out front though

EDIT: wait, is te du value you are talking about the constant? if so, i recommend moving it in front, you don't have to though. as you see, you got the right answer

6. ## Re:

Ohhh...my bad I guess what I was referring to was just the $\displaystyle \frac{1}{2}$ not the du. The du would stay behind but

7. Originally Posted by qbkr21
Ohhh...my bad I guess what I was referring to was just the $\displaystyle \frac{1}{2}$ not the du. The du would stay behind but
yes, that's what i'd do. by the way, the 4 you have in the denominator should be a 2. i'd move that out as well, so i'd have 1/4 in front