Find the integral of x/(x-1)^2 with respect to x

  • Thread starter Beretta
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  • #1
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Integral (x / (x-1)^2) dx =

Let u = x, du = dx
Let dv = 1 / (x-1)^2 , v = -1 / x-1

uv - integral (vdu) dv

x ( -1 / (x-1) ) + integral (1 / x-1 ) =

x ( -1 / (x-1) ) + ln | x - 1 |

how come the answer is ( -1 / (x-1) ) + ln | x - 1 | ? what happend to the x in ->>> x ( -1 / (x-1) ) + ln | x - 1 | ?

Thank you in advance
 

Answers and Replies

  • #2
Hurkyl
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What happened to the constant of integration?
 
  • #3
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I'm using this in differential equations I don't need the constant now.
 
  • #4
Hurkyl
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Ah, but an indefinite integral is a multivalued operation; you always need the constant of integration!

It turns out that this is problem is a good example of why it matters! When you account for the constant of integration, you can show that your answer and the "right" answer are the same!
 
  • #5
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May you show me how I can elliminate the x please?
 
  • #6
Hurkyl
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Subtract your answer from the "right" answer: what's left?
 
  • #7
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thank you SO much!
 

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