Indefinite Integrals & Substitution Rule

  • Thread starter MitsuShai
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  • #1
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Homework Statement






2. The attempt at a solution

I
 
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Answers and Replies

  • #2
Dick
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It's not even close. Since when does the integral of du/sqrt(u) have an ln in it? And shouldn't you be using du=(-5)dx to replace dx with du?
 
  • #3
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It's not even close. Since when does the integral of du/sqrt(u) have an ln in it? And shouldn't you be using du=(-5)dx to replace dx with du?
I forgot about the square root.
how about this?

I= ∫ dx/sqrt(2−5x)
= ∫ [1/sqrt(2−5x)]dx
u= 2-5x
du= -5dx
(-1/5)du = dx

∫ (-1/5)du/sqrt(u)
(-1/5) ∫ du/u^(1/2)
(-1/5) 1/ (2/3)u^(3/2)
- 1/[(10/3)u^(3/2)]
- 1/[(10/3)(2-5x)^(3/2)]
 
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  • #4
Dick
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I forgot about the square root.
how about this?

I= ∫ dx/sqrt(2−5x)
= ∫ [1/sqrt(2−5x)]dx
u= 2-5x
du= -5dx
(-1/5)du = dx

∫ (-1/5)du/sqrt(u)
(-1/5) ∫ du/u^(1/2)
(-1/5) 1/ (2/3)u^(3/2)
- 1/[(10/3)u^(3/2)]
- 1/[(10/3)(2-5x)^(3/2)]
du/sqrt(u) is u^(-1/2)du. You've still got the power rule wrong.
 
  • #5
Hurkyl
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Do remember there is a way to check your work -- what relationship is a function supposed to have with its anti-derivatives?
 
  • #6
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Do remember there is a way to check your work -- what relationship is a function supposed to have with its anti-derivatives?
Yeah, I already know that.

I don't trust my math, that's why I want to know if I'm doing this correctly.
 
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  • #7
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ok nevermind I got it.....
 

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