Integral of 1 / 1 + sqrt [x]

  • Thread starter Charas
  • Start date
  • #1
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Hi, I am a newbie here and would like to ask you stg. integral of 1 / 1 + sqrt [x]

I used chain rule and used x = u^2 is that true?

Thanx for any replies
 

Answers and Replies

  • #2
1,306
19


Do the substitutions that neutrino and morphism suggested.
 
  • #3
100
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Can't you use integration by parts?
 
  • #4


I'm doing this in my head and not fully awake, so I could be wrong.
we have 1/(1+sqrt(x)) You can use the substitution that you've shown.
u^2=x
2udu=dx
So we end up with 2u/1+u
remember you can substitute more than once....
 

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