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∫[(2)/(x-3)(√(x+10))]dx U-Substitution → du/dx or dx/du, & why?

  1. Oct 9, 2012 #1
    ∫[(2)/(x-3)(√(x+10))]dx U-Substitution → du/dx or dx/du, & why???

    Okay, here I solve [itex]\int\frac{2}{(x+3)\sqrt{x+10}}dx[/itex] in two ways.


    The problem I'm having however is during the U substitution. The first method, I take [itex]\frac{du}{dx}= [/itex] (function in terms of x)

    For the second method, I set substituted terms in terms of u and then take [itex]\frac{dx}{du}= [/itex]function in terms of u)

    So if I have indeed yielded the correct integral function using both methods, my question is is then based on my professor's suggestion that we use the second method for integrating various functions - particularly of this type with roots.

    My question is, after choosing the terms to substitute for U, how do I look at a function to determine whether to set substituted terms into terms of x or u? In other words, how do I look at a function and determine whether to use the first method du/dx or second method dx/du?

    Method 1:

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


    U substitution: Setting u equal to x terms, then du/dv

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

    [itex]\frac{du}{dx}=\frac{1}{2}(x+10)^{\frac{1}{2}-\frac{2}{2}}(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



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



    [itex]4[\frac{1}{2\sqrt7}ln|\sqrt{x+10}+\sqrt{7}|-\frac{1}{2\sqrt7}ln|\sqrt{x+10}+\sqrt{7}|]+C[/itex]

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    Method 2:

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

    let [itex]u=\sqrt{x+10}[/itex] set in terms of u

    [itex]x=u^{2}-10[/itex]

    [itex]\frac{dx}{du}=2u[/itex]

    [itex]dx=2udu[/itex]

    [itex]\int\frac{(2)(2u)du}{u^{2}-10-3u}[/itex]

    [itex]4\int\frac{u}{(u-5)(u+2)}du[/itex] And then partial fractions...

    [itex]4\int\frac{A}{(u-5)}+\frac{B}{(u+2)}du[/itex]

    ........Solving for A and B and integrating →


    [itex]4[(\frac{5}{7}ln|\sqrt{x+10}-5|)+(\frac{2}{7}ln|\sqrt{x+10}+2|)]+C[/itex]
     
    Last edited: Oct 9, 2012
  2. jcsd
  3. Oct 9, 2012 #2

    tiny-tim

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    Hi LearninDaMath! :smile:

    I'm not sure I see any difference.

    As a general principle, you should certainly get rid of all x's as soon as possible!
    shouldn't this be [itex]\int\frac{(2)(2u)du}{(u^{2}-10+3)u}[/itex] ? :wink:
     
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