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Math Help - Integration by substitution

  1. #1
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    Integration by substitution

    Hey, having a bit of trouble with a substitution problem:

    Integrate : x/(2x+1)^1/2 dx using the substitution method

    I get 1/3(2x+1)^3/2 - (2x+1)^1/2 + c

    The answer is down as 1/3(x-1)((2x+1)^1/2) + c

    I cant see how to get to this, any help would be greatly appreciated...
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  2. #2
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    What is your working? It seems sensible to pick u=2x+1 and through algebra x = (u-1)/2. From the sub we know that du = 2 dx or dx = du/2

    x/(2x+1)^(1/2) dx = (u-1)/(2u^(1/2)) * du/2 = 1/4 * [(u-1)/u^(1/2)] du

    We can simplify that up a bit to give



    Those are relatively simple integrals
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  3. #3
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    Integration by parts works nicely here...

    Let u = x so that du = dx, and let dv = (2x + 1)^(-1/2)dx so that v = (2x + 1)^(1/2)

    Then int[x(2x + 1)^(-1/2)dx] = x(2x + 1)^(1/2) - int[(2x + 1)^(1/2)dx].

    I'm sure you can go from here.
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  4. #4
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    Hello, jlaing!


    Integrate: .x/(2x+1)^{1/2} dx

    If the expression under the radical is linear, let \,u equal the entire radical.

    . . . . . . . . . _____
    We have: .√2x + 1 .= .u . , 2x + 1 .= .u^2

    . . . . . . x .= .(u^2 - 1)/2 . . dx .= .u du


    . . . . . . . . . . . . . .(u^2 - 1)/2
    Substitute: .[INT] -------------- (u du) . = . (1/2) [INT] {u^2 - 1) du
    . . . . . . . . . . . . . . . . . u

    Got it?

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  5. #5
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    Thats great thank you, it seems I can do the actual integration part but I have something basic going wrong when I am trying to get the final answer.

    For example:

    = (1/4)( (2/5)u5/2 - (2/3)u3/2 ) I can get to this line but I cant get to the next

    = ( (2x + 1)3/2(3x - 1) ) / 15 + K

    I know it is something basic but I cant see it
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  6. #6
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    you should get y= 1/6(2x+1)^1.5 - 0.5(2x+1)^0.5 + c
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  7. #7
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    Quote Originally Posted by jlaing View Post
    Thats great thank you, it seems I can do the actual integration part but I have something basic going wrong when I am trying to get the final answer.

    For example:

    = (1/4)( (2/5)u5/2 - (2/3)u3/2 ) I can get to this line but I cant get to the next

    = ( (2x + 1)3/2(3x - 1) ) / 15 + K

    I know it is something basic but I cant see it
    Use ^ to denote "to the power of" otherwise what you write is too confusing to be easily read.
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