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Math Help - Solving Integral using substitution

  1. #1
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    Solving Integral using substitution [More help needed]

    I have been having a trouble with this problem:

    X / (4X^2+1)^2 dx<-------This is an integral that I have to solve using substitution

    This is what I did.

    u=4x^2+1
    u'=8x
    du/8=dx

    So then I went back to original problem and plugged that in and got.

    1/8*(1/(1/3)u^3) I got that from taking the anti derivative of x^2
    I plugged u back in and got

    1/8 * 1/(1/3)(4x^2+1)^3

    after i simplify i get:
    1/ ((8/3)(64x^6+1))

    I would appreciate it if someone could tell me where i messed up. Thanks!
    Last edited by PensFan10; June 26th 2008 at 01:49 PM.
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  2. #2
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    Hi !

    Quote Originally Posted by PensFan10 View Post
    I would appreciate it if someone could tell me where i messed up. Thanks!
    Sure There's something wrong here :

    u=4x^2+1
    u'=8x
    du/8=dx
    Basically, u'=\frac{du}{dx}.

    So du=8xdx \implies dx=\frac{du}{8x}

    The x in the denominator will simplify in the integral with the remaining x

    \int \frac{x}{(4x^2+1)^2} dx=\int \frac{x}{u^2} \frac{du}{8x}=\frac 18 \int \frac{1}{u^2} du=\frac 18 \int u^{-2} du


    Note that the integral \int x^n dx=\frac{x^{n+1}}{n+1}.

    Here, n=-2 --> n=-1, not -3.
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  3. #3
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    okay

    Thanks. After I plug u back in I still get the wrong answer. If it's not to much of a hassle, could you finish the problem for me?
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  4. #4
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    okay

    I got that far. I then took the anti derivative of u^-2 and got -U^-1

    I plugged u into that equation and got -((4x^2)+1)^(-1+2) = -((4x^2)+1)+c Then I multiply that by 1/8 which gives:

    -((4x^2)+1)+c / 8

    I must be messing up with the algebra part because I kept getting the same thing you posted.

    My teacher gave -1/(32x^2+8) + c as the correct answer.
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  5. #5
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    Quote Originally Posted by PensFan10 View Post
    I got that far. I then took the anti derivative of u^-2 and got -U^-1

    I plugged u into that equation and got -((4x^2)+1)^(-1+2) = -((4x^2)+1)+c Then I multiply that by 1/8 which gives:

    -((4x^2)+1)+c / 8

    I must be messing up with the algebra part because I kept getting the same thing you posted.

    My teacher gave -1/(32x^2+8) + c as the correct answer.
    Hey ! Yes, you are messing up with the algebra (and the formula ?).

    You have \frac 18 \int u^{-2} du=\frac 18 (-u^{-1})=-\frac 18 \cdot \frac 1u=-\frac{1}{8(4x^2+1)}.

    Better ?
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  6. #6
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    Awesome!

    Yeah. That is much better. I should have just made it 1/u rather than trying to mess with the exponents. It all makes sense now
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