U-substitution integral

  • Thread starter mmapcpro
  • Start date
  • #1
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I'm trying to re-study the calculus I learned years ago to resume my degree this fall, so I don't have access to an instructor right now. I'm stuck on a problem that I can't figure out, and it's just bugging me.

Homework Statement



[tex]\int[/tex] 1/(1 + [tex]\sqrt{2x}[/tex]) dx


Homework Equations


I'm trying to use u-substitution, but I am not seeing it.


The Attempt at a Solution



I have thought to rationalize the denominator first, which would allow me to split it up into 2 integrals (1/1-2x) and ([tex]\sqrt{2x}[/tex]/(1-2x) but I get stuck
 
Last edited:

Answers and Replies

  • #2
Dick
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Try u=1+sqrt(2x). What's dx in term of u?
 
  • #3
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I set u = 1 + [tex]\sqrt{2x}[/tex], so then u-1 = [tex]\sqrt{2x}[/tex], so 2x = (u -1)[tex]^{2}[/tex]

so then x = 1/2(u - 1)[tex]^{2}[/tex]

then dx = (u - 1)du

so I could set it up as [tex]\int[/tex] (u - 1)du / u

split that up into [tex]\int[/tex] 1du - [tex]\int[/tex] du/u

that leaves me with u - ln|u| + c

making it 1 + [tex]\sqrt{2x}[/tex] - ln|1+[tex]\sqrt{2x}[/tex]| + c

But the solution in the book reads: [tex]\sqrt{2x}[/tex] - ln|1 + [tex]\sqrt{2x}[/tex]| + c

I feel like an idiot...it's been about 10 years since I've done this.
 
  • #4
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Does the +1 simply get absorbed into the constant, C?
 
  • #5
SammyS
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Does the +1 simply get absorbed into the constant, C?
Yes.
 

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