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Hard Antiderivative

  1. Mar 10, 2013 #1
    1. The problem, the whole problem, and nothing but the problem

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



    2. Relevant equations

    u-substitution (in the style of trig substitution)
    I think that I've got it figured out, I just don't know if my substitutions were legitimate.


    3. The attempt at a solution

    [tex] \int \frac{dx}{ \sqrt{ 1+ \sqrt{ 1+ \sqrt{ x } } } } [/tex]
    [tex] u^2 = x, 2u \, du = dx [/tex]
    [tex] 2 \int \frac{u}{ \sqrt{ 1+ \sqrt{ 1+ u } } } du [/tex]
    [tex] v^2 = 1+u, 2v \, dv = du [/tex]
    [tex] 4 \int \frac{v(v^2 - 1)}{ \sqrt{ 1+ v } } dv [/tex]

    Now just simplifying:

    [tex] 4 \int \frac{v(v+1)(v-1)\sqrt{ 1+ v }}{ v+1 } dv [/tex]
    [tex] 4 \int v(v-1)\sqrt{ 1+ v } \, dv [/tex]
    [tex] w^2 = 1+v, 2w \, dw = dv [/tex]
    [tex] 8 \int w(w^2 -1 )(w^2 - 2)(w) \, dw [/tex]
    [tex] 8 \int w^2(w^2 -1 )(w^2 - 2) \, dw [/tex]

    I'm sure that I could foil this out, integrate, and make the 4 or 5 substitutions. Before I do so, is my work justified? If so, is there a trick to the last integral (without foiling it all out)?

    Thanks
     
  2. jcsd
  3. Mar 10, 2013 #2
    If you foil it out it becomes w6-3w4+2w2... You can solve this integral pretty easily I assume?
    And it looks good to me
     
  4. Mar 10, 2013 #3
    Oh, that's embarrassing, I thought that it was going to be ugly...

    [tex] = 8 \int w^6 - 3w^4 + 2w^2 dw [/tex]
    [tex] = 8 \big[ w^7/7 -3w^5/5 + 2w^3/3 \big] [/tex]
    [tex] =8 \big[ (1+v)^{7/2}/7 -3(1+v)^{5/2}/5 + 2(1+v)^{3/2}/3 \big] [/tex]
    [tex] =8 \big[ (1+\sqrt{1+u})^{7/2}/7 - 3(1+\sqrt{1+u})^{5/2}/5 + 2(1+\sqrt{1+u})^{3/2}/3 \big] [/tex]
    [tex] =8 \big[ \big(1+\sqrt{1+\sqrt{x}}\big)^{7/2}/7 - 3\big(1+\sqrt{1+\sqrt{x}}\big)^{5/2}/5 + 2\big(1+\sqrt{1+\sqrt{x}}\big)^{3/2}/3 \big] [/tex]

    [tex] = \frac{ 8\big(1+\sqrt{1+\sqrt{x}}\big)^{7/2}}{7} - \frac{24\big(1+\sqrt{1+\sqrt{x}}\big)^{5/2}}{5} + \frac{16\big(1+\sqrt{1+\sqrt{x}}\big)^{3/2}}{3} [/tex]
     
  5. Mar 10, 2013 #4

    Ray Vickson

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    I don't know what "foiling out" means, but the final integral is easy.
     
  6. Mar 10, 2013 #5

    SammyS

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    Ray,

    FOIL is a mnemonic for multiplying (expanding) two binomials.

    The product of the First term of each binomial
     plus
    the product of the two Outer terms
     plus
    the product of the two Inner terms
     plus
    the product of the Last term of each binomial .

    [itex]\displaystyle (a+b)(c+d) = \underbrace{ac}_\mathrm{first} + \underbrace{ad}_\mathrm{outside} + \underbrace{bc}_\mathrm{inside} + \underbrace{bd}_\mathrm{last}[/itex]


    The terminology has come to be used in a more general (and incorrect) way to refer to expanding the product of two or more polynomials.
     
  7. Mar 10, 2013 #6

    SammyS

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    At this point, the substitution
    w = 1+v ​
    would work well.

    It gives [itex]\displaystyle \int (w-1)(w-2)(w^{1/2})\, dw\ .[/itex]
     
  8. Mar 10, 2013 #7
    I got the problem a while ago, Samuel.∇
     
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