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Integration math question

  • Thread starter Gear300
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  • #1
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If we were to integrate 1/x by parts using u = 1/x and dv = 1dx, then it would end up as:
Int(dx/x) = 1 + Int(dx/x), ending up with 1 = 0. Why was there a discrepancy in the integration?
--Int() refers to integral
 

Answers and Replies

  • #2
morphism
Science Advisor
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You're missing an integration constant.
 
  • #3
exk
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[tex]\int\frac{1}{x}dx=uv -\int vdu= \frac{x+c}{x} - \int \frac{-(x+c)}{x^{2}}dx = etc...[/tex]
[tex]u=\frac{1}{x}[/tex]
[tex]du=\frac{-1}{x^{2}}[/tex]
[tex]dv=1dx[/tex]
[tex]v=x+c[/tex]

Please tell me if I am wrong.

Regardless, you are looking for the result to be ln|x|+c.
 
  • #4
Defennder
Homework Helper
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[tex]v=x+c[/tex]
I don't think there should be a '+ c' in this intermediate step. Anyway, integrating
1/x gives ln(x) + c by definition. Some books actually define ln(x) to be the anti-derivative of that.
 

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