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help = integration problem! (1 Viewer)

asb_92

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hey guys
ive gotta mental blank...:confused:
how do u integrate:

1/(3-x)^2

thanks heaps for the help!! :)
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cutemouse

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Integral is: sin-1(x/sqrt(3)) + c

EDIT: Wrong answer.. whoops =P
 
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Fortify

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if its: 1/(3-x^2)

1/sqrt3 tan^-1 (x/sqrt3) + c
 
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lychnobity

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Hang on a minute... isn't it 1/(3 - x) + C ??
 
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asb_92

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ooo ok
thanks heaps!! :D
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Fortify

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asb92, do you mean:

1/(3-x)^2

or

1/(3-x^2) ?
 

kwabon

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hey guys
ive gotta mental blank...:confused:
how do u integrate:

1/(3-x)^2

thanks heaps for the help!! :)
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integral of (3-X)^(-2)
[ (3-X)^(-1)/(-1 * -1) ]
therefore = 1/(3-X) + c
 
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Fortify

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If it's 1/(3-x)^2

Then the answer is 1/(3-x) + C
 

lychnobity

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For the integral to be an inverse sign, the question would need a square root sign over the whole of the denominator...
 

Fortify

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i srsly dont see how you got sine inverse, even if it was 1/(3-x^2), how is that a sine inverse???
i am pretty sure thats 4 unit and u will have to use partial fractions.
sorry DAI LO .

I misread it ok ? chill cuz, im not a robot, i make mistakes ok ?

If its 1/(3-x^2)

then its

1/sqrt3 tan^-1 (x/sqrt3) + c
 

cutemouse

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Sorry, I misread the question... Disregard my answer =)
 

Fortify

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and does my reply really sound that arrogant. if so sorry....:)
Sorry man, I wasn't in the mood before. I feeling better now. I was distracted by something else on my mind.
 

asb_92

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lol im confused... :S

umm its 1/(3-x)^2


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