Question

Integration.
Spot any errors?
Thanks in advance.

Answer 1

i think this is right, let me do it another way and see if we get the same answer
i would put \(u=2x+1, x=\frac{u}{2}-\frac{1}{2}, du= 2dx\)

Answer 2

oh and so \(\frac{1}{2}du = dx \)

Answer 3

get
\[\frac{1}{2}\int\frac{u}{2\sqrt{u}}du-\frac{1}{2}\int\frac{1}{2\sqrt{u}}du\] or maybe factor out the \(\frac{1}{2}\)

Answer 4

then rewrite with exponents and integrate that way

Answer 5

then i substitute sqrt(2x+1) , right?

Answer 6

looks like we get the same answer though, so no problem here

Answer 7

yeah after integrating

Answer 8

Wait. why are you getting a negative sign in between?

Answer 9

we get the same thing as an answer
this method gives \(\frac{1}{6}u^{\frac{3}{2}}-\frac{1}{2}u^{\frac{1}{2}}\)

Answer 10

But i don't get the format of wolfram. :S

Answer 11

i get a negative sign because if i make \(u=2x+1\) then the \(x\) in the numerator is \(\frac{u}{2}-\frac{1}{2}\)

Answer 12

http://www.wolframalpha.com/input/?i=integrate+x%2Fsqrt(2x%2B1)+dx%2C+2x%2B1%3Dt%5E2

Answer 13

how you got this step?
\[\frac{1}{2}\int\limits\frac{u}{2\sqrt{u}}du-\frac{1}{2}\int\limits\frac{1}{2\sqrt{u}}du \]

Answer 14

maybe my constants are off
if you look here
http://www.wolframalpha.com/input/?i=integal+x%2Fsqrt%281%2B2x%29
wolf does it the same way i did

Answer 15

i think maybe you are missing this:
if you make the u - sub \(u=2x+1\) you have to solve for \(x\) for the \(x\) in the numerator

Answer 16

In your way you found the substitution thing on your own, right?
In our syllabus, the substitution thing will always be given.

Answer 17

once you do that, you get
\[x=\frac{1}{2}u-\frac{1}{2}\] so you break the integral in to two parts

Answer 18

Sorry. Idk what sub u is. :S

Answer 19

it is the chain rule in reverse basically

Answer 20

Now i get what you did in the last step^

Answer 21

you have a composition of functions and you put \(u\)= the inside piece

Answer 22

i guess we can do it your way too

Answer 23

And how you got sqrtu in the denominators?

Answer 24

So is my answer right?

Answer 25

yes, now you get \(\sqrt{u}\) in the denominator, so you can use the power rule rule

Answer 26

you are off by a constant somewhere let me find it

Answer 27

Thank you!

Answer 28

well no actually you are not. your answer is correct

Answer 29

And i still don't get how you got sqrt u in the denominator?!

Answer 30

Yay! :D

Answer 31

oh because i turned \(u\) in to \(2x+1\) and so \(\sqrt{2x+1}=\sqrt{u}\)

Answer 32

Shouldn't that go in the numerator?

Answer 33

no because it is in the denominator

Answer 34

here is a simpler example
if i had
\[\int\frac{x^2dx}{\sqrt{x^3+5}}\] would make \(u=x^3+5,du=3x^2dx,\frac{1}{3}du=x^2dx\) and rewrite the whole thing as
\[\frac{1}{3}\int\frac{du}{\sqrt{u}}\]

Answer 35

yeah, that is the numerator, replace \(x\) by \(\frac{u}{2}-\frac{1}{2}\) but the denominator is still \(\sqrt{u}\)

Answer 36

http://static2.fjcdn.com/comments/i+see+what+you+did+there+_d5597f17acded4c87204270fbe524d9e.jpeg

Answer 37

Thank you! :D

Answer 38

lol
yw but you have it right, so good work

Answer 39

Thank you.