text: # lnx Integral
  
  • Dec 9th 2007, 08:20 AM
  akhayoon
  lnx Integral
  $\int \frac{ln|x|} {x\sqrt{1+ln|x|}}$?
  • Dec 9th 2007, 08:23 AM
  ThePerfectHacker
  Let $t=\ln x$ ---> trivial excercise.
  • Dec 9th 2007, 08:27 AM
  akhayoon
  I don't get it
  • Dec 9th 2007, 10:40 AM
  Krizalid
  If $t=\ln x\implies dt=\frac1x\,dx.$
  
  It's just substitution.
  
  Can you handle it from there?
  • Dec 9th 2007, 11:26 AM
  akhayoon
  but then I get
  
  $\int\frac{u}{\sqrt{1+u}}~dt$
  
  how do I integrate this?
  • Dec 9th 2007, 11:31 AM
  Krizalid
  Substitute $\varphi ^2 = 1 + u,$
  
  $\int {\frac{u}
  {{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .$
  • Dec 9th 2007, 11:42 AM
  akhayoon
  Quote:
  
  Originally Posted by Krizalid
  Substitute $\varphi ^2 = 1 + u,$
  
  $\int {\frac{u}
  {{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .$
  
  $\int\frac{\varphi^{2}-1}{\varphi}~d\varphi$
  
  so to get ur answer we had to turn $\frac{1}{\varphi}$ into 2?
  
  how does that work
  
  I'm sorry but I'm just trying to understand the solution(Angel)
  • Dec 9th 2007, 12:14 PM
  Jhevon
  Quote:
  
  Originally Posted by akhayoon
  $\int\frac{\varphi^{2}-1}{\varphi}~d\varphi$
  
  so to get ur answer we had to turn $\frac{1}{\varphi}$ into 2?
  
  how does that work
  
  I'm sorry but I'm just trying to understand the solution(Angel)
  
  Krizalid thinks a few levels above the average person when it comes to integrals. here is a more conventional substitution for mortals.
  
  $\int \frac u{\sqrt{1 + u}}~du$
  
  Let $t = 1 + u \implies \boxed{u = t - 1}$
  
  $\Rightarrow dt = du$
  
  So our integral becomes:
  
  $\int \frac {t - 1}{\sqrt{t}}~dt$
  
  which we can do using the power rule (after dividing $\sqrt{t}$ into each of the terms in the numerator)
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distilabel_metadata.content: # lnx Integral
  
  • Dec 9th 2007, 08:20 AM
  akhayoon
  lnx Integral
  $\int \frac{ln|x|} {x\sqrt{1+ln|x|}}$?
  • Dec 9th 2007, 08:23 AM
  ThePerfectHacker
  Let $t=\ln x$ ---> trivial excercise.
  • Dec 9th 2007, 08:27 AM
  akhayoon
  I don't get it
  • Dec 9th 2007, 10:40 AM
  Krizalid
  If $t=\ln x\implies dt=\frac1x\,dx.$
  
  It's just substitution.
  
  Can you handle it from there?
  • Dec 9th 2007, 11:26 AM
  akhayoon
  but then I get
  
  $\int\frac{u}{\sqrt{1+u}}~dt$
  
  how do I integrate this?
  • Dec 9th 2007, 11:31 AM
  Krizalid
  Substitute $\varphi ^2 = 1 + u,$
  
  $\int {\frac{u}
  {{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .$
  • Dec 9th 2007, 11:42 AM
  akhayoon
  Quote:
  
  Originally Posted by Krizalid
  Substitute $\varphi ^2 = 1 + u,$
  
  $\int {\frac{u}
  {{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .$
  
  $\int\frac{\varphi^{2}-1}{\varphi}~d\varphi$
  
  so to get ur answer we had to turn $\frac{1}{\varphi}$ into 2?
  
  how does that work
  
  I'm sorry but I'm just trying to understand the solution(Angel)
  • Dec 9th 2007, 12:14 PM
  Jhevon
  Quote:
  
  Originally Posted by akhayoon
  $\int\frac{\varphi^{2}-1}{\varphi}~d\varphi$
  
  so to get ur answer we had to turn $\frac{1}{\varphi}$ into 2?
  
  how does that work
  
  I'm sorry but I'm just trying to understand the solution(Angel)
  
  Krizalid thinks a few levels above the average person when it comes to integrals. here is a more conventional substitution for mortals.
  
  $\int \frac u{\sqrt{1 + u}}~du$
  
  Let $t = 1 + u \implies \boxed{u = t - 1}$
  
  $\Rightarrow dt = du$
  
  So our integral becomes:
  
  $\int \frac {t - 1}{\sqrt{t}}~dt$
  
  which we can do using the power rule (after dividing $\sqrt{t}$ into each of the terms in the numerator)
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a series of posts from a math forum where a user asks for help with integrating a specific function, and other users provide guidance and solutions. The initial post contains a math question, and subsequent posts contain answers and further explanations.
  
  Label: question-and-answer