Question

integral of x/((3-x^2) ^2)

Answer 1

let u= \(3 - x^2 \rightarrow du = -2xdx\)

Answer 2

therefore you have \[\frac 12 \int \frac{du}{u^2}\]

Answer 3

I got \[\int\limits (x) * (3-x^2) ^{-1/2} dx\]

u = 3-x^2
du = 2x dx

1/2du = xdx

\[1/2 \int\limits (u)^{-1/2} + c\]

\[(1/2) * (-2/3) u^{-2/3}\]

Which should equal\[-{2/6} (3 - x^2)^ {-3/2}\]

Answer 4

waut...why -1/2??

Answer 5

isnt that squared only?

Answer 6

At first I thought it wasn't negative, but then I thought it does that when it goes from the denom to the num?

Answer 7

i meant why 1/2

Answer 8

the denominator is squared right?

Answer 9

I'm sorry it was supposed to be \[\sqrt{3-x^2}\]

Answer 10

I messed up :(

Answer 11

your process is all right except the final answer

Answer 12

1) simplify 2/6 into 1/3

2) where did 3/2 come from?

Answer 13

oh you miswrote the previous step as 2/3 instead of 3/2

Answer 14

nevermind 3/2 is correct

Answer 15

Yeah I messed up, I had different stuff written at first and messed up everything else sorry.

Answer 16

in simplified term it should look like \[-\frac{1}{3(3-x^2)^{3/2}}\]

Answer 17

now that I think about it shouldn't it be -1/2 + 2/2 to be Positive 1/2?

Answer 18

good point

Answer 19

how could i have overlooked that lol

Answer 20

so 1/2 cancels leaving you with \((3-x^2)^{1/2}\) only

Answer 21

so it should be \[-(1/3) * (3 - x^2)^{1/2} + C\]

Answer 22

-1/3? no

Answer 23

integral of (3-x^2)^-1/2 is \(2(3 - x^2)^{1/2}\) agree?

Answer 24

oh yeah woopsies, keep confusing myself lol. I just looked back at your last post.

Answer 25

so then there's a 1/2 in the outside..2 just cancels

Answer 26

\[(3−x^2)^{1/2}+C\]

Answer 27

^that

Answer 28

apparently that isn't correct, I'm going to try this over from scratch.

Answer 29

no?!

Answer 30

lemme clarify this...this is the question? \[\int \frac{x}{\sqrt{3-x^2}}\]

Answer 31

yup

Answer 32

OHHH of course *facepalm* the negative sign!!

Answer 33

it's \[-\frac{1}{2} [2(3-x^2)^{1/2}]\] there should be a negative sign

Answer 34

sorry for overlooking a lot of things :S

Answer 35

oopsies :P No it's okay, I appreciate all of your help :).

Answer 36

I have messed up badly on this question :P
so \[-(3-x^2) ^{1/2}\]

Answer 37

+C :P lol

Answer 38

That's an obvious :P. By the way when do we actually find what C is? I feel like it's always something we write, but never figure out...

Answer 39

Now that I think about it how did we get -1/2? Was it because 3-x^2 was -2x dx?

Answer 40

That would make sense I think I overlooked it :)

Answer 41

yes derivative of 3-x^2 is -2xdx

Answer 42

and you will know what C is if there are numbers on the integral like \[\int_2^1 x^2dx\] don't mind that for now lol

Answer 43

Ah okay, I've dealt with those already, I figured there was somehing else to it, thanks ! :).