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Homework Help: Is this an acceptable route to take for solving this integral involving roots:

  1. Oct 8, 2012 #1

    First thing would be u-substitution, finding what I can replace in terms of u:

    let [itex]u=\sqrt{x+10}[/itex]


    [itex]du=\frac{1}{2\sqrt{x+10}}dx[/itex] → [itex]dx=2\sqrt{x+10}du[/itex]

    Then replace those substitutions into the integral and simplify what I can:

    [itex]\int\frac{(2)(2)\sqrt{x+10}du}{(x+3)\sqrt{x+10}}[/itex] → [itex]4\int\frac{du}{(x+3)}[/itex]

    Then realize I still have an x term and figure out a way to arrange the previous substitutions to eliminate the remaining x term

    [itex][u^{2}-10=x][/itex] → [itex][u^{2}-10+3=x+3][/itex] → [itex][u^{2}-7=x+3][/itex]

    [itex]4\int\frac{1}{u^{2}-7}du[/itex] .....And then partial fractions from here

    Is this a correct track so far?
  2. jcsd
  3. Oct 8, 2012 #2
    Looks good.
  4. Oct 8, 2012 #3
    Cool, thanks, so in proceeding into partial fractions, I am a little hesitant in how to set it up. I have two ways in mind.

    Attempt 1:

    4[itex]\int\frac{1}{u^{2}-7}[/itex] = [itex]\frac{Au+B}{u^{2}-7}[/itex]

    [1 = Au + B], ? Maybe B = 1, but then what does A equal? Would A=0?

    Setting up like this doesn't feel right, but I'm not exactly sure why.
    Reason1: I think the goal of partial fractions is to turn 1 fraction into more than one fraction.
    Reason2: Setting the numerators equal to each other in this configuration does not allow for solving for u.
    Attempt 2:

    4[itex]\int\frac{1}{u^{2}-7}[/itex] = [itex]\frac{A}{u+\sqrt{7}}+\frac{B}{u-\sqrt{7}}[/itex]

    Which becomes: [itex]1=A(u-\sqrt{7})+B(u+\sqrt{7})[/itex] → 1 = u(A+B) + ([itex]\sqrt{7}-\sqrt{7}[/itex] →

    1 = u(A+B) ....which doesn't seem like something that can be solved for A or B.

    Which of these methods would be correct or incorrect?
    Last edited: Oct 8, 2012
  5. Oct 8, 2012 #4
    The second method you used is correct: partial fractions is used to split a rational function, one that cannot be simplified by dividing, up into simpler rational functions with only linear or unfactorable quadratic terms in the denominators. These terms are the factors of the original denominator.
    To solve your equation for A and B separately, you must realize you actually have a system of an infinite amount of equations. Since all you did was break the fraction into additive components, the equation must be true for every value of u. Simply choose a value of u that removes all of the variables but one. For example, you can choose the particular case when [itex]u = \sqrt{7}[/itex], which allows you to solve for the value of B. You can use another value of u to find an equation where B is not present and solve for the value of A.
  6. Oct 8, 2012 #5
    A(u - \sqrt 7) + B(u + \sqrt 7) = (A + B)u + (B - A)\sqrt 7 = 1
    [/tex] thus [tex]
    A + B = 0
    \\ (B - A)\sqrt 7 = 1
  7. Oct 9, 2012 #6
    Thanks, from there, I have tried to work it out, however it just doesn't feel correct for some reason. These are all the steps I made, is it correct?

    [itex]4\int\frac{1}{u^{2}-7}[/itex] = [itex]\frac{A}{u+\sqrt{7}}+\frac{B}{u-\sqrt{7}}[/itex]

    A(u - \sqrt 7) + B(u + \sqrt 7) = (A + B)u + (B - A)\sqrt 7 = 1
    [/tex] thus [tex]
    A + B = 0
    \\ (B - A)\sqrt 7 = 1

    [tex] [A + B = 0]
    \\ \sqrt 7B - \sqrt 7A = 1

    [tex] [\sqrt7B + \sqrt 7A = 0]
    \\ \sqrt 7B - \sqrt 7A = 1[/tex] (subtracting to solve for B)

    [itex]2\sqrt{7}A=-1[/itex] → [itex]A=-\frac{1}{2\sqrt{7}}[/itex]

    Subbing A into [itex][A + B = 0] → B = \frac{1}{2\sqrt{7}}[/itex]


    [itex]4\int\frac{1}{u^{2}-7}[/itex] = [itex]\frac{A}{u+\sqrt{7}}+\frac{B}{u-\sqrt{7}}[/itex] becomes

    [itex]4\int\frac{1}{u^{2}-7}= \int\frac{\frac{1}{2\sqrt7}}{u+\sqrt{7}}+\int\frac{-\frac{1}{2\sqrt7}}{u-\sqrt{7}}[/itex]

    [itex]4\int\frac{1}{u^{2}-7}= \frac{1}{2\sqrt7}\int\frac{1}{u+\sqrt{7}}-\frac{1}{2\sqrt7}\int\frac{1}{u-\sqrt{7}}[/itex]


  8. Oct 9, 2012 #7
    You computed A and B correctly, but when you plugged them into the integral you swapped their signs for some reason. Apart from this (and the subsequent error) everything seems OK.
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