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A few more Integrals

  1. Apr 21, 2006 #1
    Find the integral.


    This one is just so confusing. *sigh*

    Find the antiderivative.


    I don't know how to approach this one.. I'm guessing making some substitutions.. but i don't know how you actually work it when you make subtitutions.. just like in my other post.. I can make substitutions.. but i don't know what to do after that. I need one with substitutions worked for me if they're all similar.
  2. jcsd
  3. Apr 21, 2006 #2


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    Would it help you to see how a substitution works if I worked out a simple problem?

    [tex]\int\frac{2x}{\sqrt{x^2+1}} \ dx[/tex]


    [tex]\int\frac{1}{\sqrt{u}} \ du[/tex] <-Substitute values for u and du as appropriate.

    [tex]\int u^{-\frac{1}{2}} du[/tex]<-Just rewriting the square root sign as a power of -1/2 to make it easier to see the integration.


    [tex]2(x^2+1)^{\frac{1}{2}}[/tex]<--substitute back for u=x2


    Take a minute to understand why the substitution worked. You want to put everything in terms of one variable. By choosing u to be the value in the square root you obtain a value of du that matches the other x and dx values.
    Last edited: Apr 21, 2006
  4. Apr 21, 2006 #3
    How's this for the first one?


    And this for the second?


    Another one came out with an undefined answer, but the answer in the back of the book turned it into natural logs.. hmm.. i don't know how that works.. But following the same pattern.. i'd get..


  5. Apr 21, 2006 #4


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    Well see du is the differential of u...

    If u is [tex]x^2+2x+2[/tex] then du would be [tex](2x+2)dx[/tex]
    Last edited: Apr 21, 2006
  6. Apr 21, 2006 #5
    lol forget i said that..

    ok, i understand how to get what du is equal to.. what i don't get now is.. in your example problem, you never substituted back for the value of du.. so would it even change my answer?
    Last edited: Apr 21, 2006
  7. Apr 22, 2006 #6
    In the second integral: u=1-4x so du=-4dx NOT 2dx.
    and [tex] \int u^{-1} du = lnu NOT u^{-1}/0 [/tex]
    Last edited: Apr 22, 2006
  8. Apr 22, 2006 #7
    Look a bit more carefully at dav2008's example. Do you see du anywhere in the expression: 2u1/2? That's why he never substituted the value of du back in.
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