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Homework Help: A few more Integrals

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

    http://img83.imageshack.us/img83/1228/int203bd.gif [Broken]

    This one is just so confusing. *sigh*

    Find the antiderivative.

    http://img83.imageshack.us/img83/7179/int264ol.gif [Broken]

    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.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 21, 2006 #2


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    Gold Member

    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?

    http://img97.imageshack.us/img97/7520/inttry206vz.gif [Broken]

    And this for the second?

    http://img186.imageshack.us/img186/6919/noworkint266iv.gif [Broken]

    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..

    http://img88.imageshack.us/img88/1598/workint262dh.gif [Broken]

    Last edited by a moderator: May 2, 2017
  5. Apr 21, 2006 #4


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    Gold Member

    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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