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Homework Help: Integrate this one?

  1. Apr 14, 2010 #1
    how am i going to integrate this one?

    f(x) = sqrt( 1 + 1/x )

    sorry for not using the symbol.. because not so familiar with that .. ^.^
     
  2. jcsd
  3. Apr 14, 2010 #2
    try substituting tan u = x
     
  4. Apr 14, 2010 #3
    does tan u = x work with f(x) = sqrt( 1 + (e^x)/4 ) ?
     
  5. Apr 14, 2010 #4
    did you solve the earlier?

    it didn't work

    if tan u = (e^(x/2)) / 2 would work i guess
     
  6. Apr 14, 2010 #5
    waaaaaaa, how come, i get sqrt(4+ex)

    thats wierd
     
    Last edited: Apr 14, 2010
  7. Apr 14, 2010 #6
    how did you figure out ? By intuition ?
    sorry , i think i have some mistake on the earlier question i posted .

    i forgot to raise the power of x to 2
    the f(x) should be f(x) = sqrt( 1 + 1/(x^2) )

    sorry about that ..~
     
  8. Apr 14, 2010 #7

    gabbagabbahey

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    Have you figured this one out yet?
     
  9. Apr 14, 2010 #8
    owho, if your question sqrt( 1 + 1/x ) then use tan u = sqrt(x)


    if sqrt( 1 + 1/(x^2) ) then use tan u = x


    hmm, there are many ways, but for me, i imagine of a triangle

    sqrt( 1 + 1/(x2) ) = (sqrt(x2+1))/ x

    hmm how should i draw the triangle,

    or wait until some other explanation people give,

    i try scanning my triangle for a while. ngahahaha
     
  10. Apr 14, 2010 #9
    View attachment scan0001.pdf

    there, hmm how do i choose those values on triangle by intuition i guess, try to make it same as your equation.

    im really good in english tough, i hope someone can explain it more detail
     
  11. Apr 14, 2010 #10
    haven't ..~ no idea before annoymage replied my post~

    annoymage : woww..~ never thought of that methods ...~
     
    Last edited: Apr 14, 2010
  12. Apr 14, 2010 #11

    gabbagabbahey

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    In that case, you'll probably find it easiest if you make the substitution [itex]w=\frac{1}{x}[/itex] and integrate by parts once before making the trig substitution annoymage suggested. Give it a try and post your attempt.
     
  13. Apr 15, 2010 #12
    I don't think integration by parts will go anywhere.

    Integration tables.

    You can obtain the the first integration form by substitution.

    [tex]\int\frac{\sqrt{a+bu}}{u^2}du=\frac{-\sqrt{a+bu}}{u}+\frac{b}{2}*\int\frac{du}{u\sqrt{a+bu}}[/tex]

    [tex]\int\frac{du}{u\sqrt{a+bu}}=\frac{1}{\sqrt{a}}*ln{\left|\frac{\sqrt{a+bu}-\sqrt{a}}{\sqrt{a+bu}+\sqrt{a}}\right|}[/tex]
     
  14. Apr 16, 2010 #13

    gabbagabbahey

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    This is integration by parts. The tables of integrals are derived using the same methods of calculus taught to students.
     
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