Question

Please help! i have solution and answer but i still dont understand how they got to the answer.

Answer 1

\[\int\limits_{}^{}\sqrt{2x+1}dx\]

Answer 2

let
u=2x+1 du=2dx dx=1/2

\[\int\limits_{}^{}\sqrt{u}*1/2du=1/2\int\limits_{}^{}u^{1/2}du \]
\[=1/2*u^{3/2}/(3/2)+C\]
=1/3(2x+1)6{3/2}+C

Answer 3

why its sqrt u?? in the dirst place

Answer 4

If u=2x +1, then the expression under the radical is u. So, sq rt of u is the same thing as u^(1/2).

Answer 5

is it beacuse sqrt(2x+1)?

Answer 6

kapeesh?

Answer 7

whats that ?

Answer 8

lol....too clear or not clear enough?

Answer 9

@Hoa, no problem, any help from friends is always appreciated.

Answer 10

teamwork play ! anyway, so why its sqrt (u)?

Answer 11

then it says 2/2 du.

Answer 12

1/2

Answer 13

Sq rt is the same thing as the power 1/2

Answer 14

I see. then = 1/2 in the front of sign of S u1/2 du. where did sqrt u go?

Answer 15

Since sq rt is the same thing as 1/2, they change \[\int\limits_{}^{}\sqrt{u}du=\int\limits_{}^{}u ^{\frac{ 1 }{ 2 }}du\]

Answer 16

The 1/2 in front of the integral is because u = 2x +1 and du=2dx, which means that xdx =(1/2)du. So the 1/2 just goes out in front of the integral.

Answer 17

I see. these questions are really confusing. i will open open up the new question. would you help me with that?

Answer 18

I have time for one more question. post it.

Answer 19

thank you so much! k