Answer to \ 2/(x-2)^3 dx

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  • #1
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Homework Statement


\ 2/(x-2)^3 dx
Basically integrating a perfect cube in the denominator with a constant in the numerator


Homework Equations





The Attempt at a Solution


i thought it would be a form of ln(x), but then, that would mean having atleast some x terms in the numerator which are not there, so, how do i do this? Is there a known pre-fixed solution for these things? Like exp(something)?
 

Answers and Replies

  • #2
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Use [tex]\int x^ndx = \frac{x^{n+1}}{n+1} + c[/tex]
 
Last edited:
  • #3
cristo
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Try writing as 2(x-2)-3. Do you know how to integrate x-3?
 
  • #4
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Thanks Arun and Cristo.
Cristo, I dont know how to integrate x^-3. (i think i might be hopeless, right?)
Arun, what is the expansion formula you just gave me called? Is there a name for solving by that method? It is applicable to negative powers as well?
 
  • #5
cristo
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Arunbg's formula is the formula for integrating a polynomial function of x. Thus, to integrate x-3 you would use that formula.
 
  • #6
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you want to integrate 2(x-2)-3

using this [tex]\int x^n = \frac{x^{n+1}}{n+1} + c[/tex]

the first thing you notice is that you add 1 to the power, then you divided by the new power
after that you multiply by the differntial of what is inside the brackets
 
  • #7
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in your question the power is -3
 
  • #8
cristo
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you want to integrate 2(x-2)-3

using this [tex]\int x^n = \frac{x^{n+1}}{n+1} + c[/tex]

the first thing you notice is that you add 1 to the power, then you divided by the new power
after that you multiply by the differntial of what is inside the brackets

The last bit should read: you divide by the derivative of the term inside the brackets.

(I know it's probably a typo, and it doesn't matter in this case; but it may confuse the OP in future if left uncorrected)
 
  • #9
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yes definitely a typo :blushing:
you always divid by the differential of the inside of the brackets when integrating!
 

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