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Math Help - Another integral.

  1. #1
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    Red face Another integral.

    Okay,
    Can I write like this:
    integral x^-4=x^-3/-3 ?
    And one more question, from where is 8/9? In this exercise
    Last edited by Kristina; November 14th 2010 at 09:12 AM.
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  2. #2
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    Quote Originally Posted by Kristina View Post
    Can I write like this:
    integral x^-4=x^-3/-3 ?
    Yes, \displaystyle \int{\frac{1}{x^4}\;{dx} = -\frac{1}{3x^3}+k.
    And one more question, from where is 8/9? In this exercise
    After differentiating, solve for [LaTeX ERROR: Convert failed] and multiply both sides by 8x^2:

    \displaystyle u = 3x^3-1 \Rightarrow \frac{du}{dx} = 9x^2 \Rightarrow dx = \frac{1}{9x^2}\;{du} \Leftrightarrow 8x^2dx = \frac{8}{9}\;{du}.
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  3. #3
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    So,
    1)-1/3x^3 fluxcion will be -3x^-3=9x^-4 I'm not right?
    2)HOwever I have integration simalar like this example and I tried to use it. But I find problem
    Integration (x^2-2) ln(x)dx
    u=x^2-2
    du=2x
    du/dx=2x
    dx=du/2x
    ln(x)/2x*du
    What i do wrong?
    Last edited by Kristina; November 15th 2010 at 05:12 AM.
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Kristina View Post
    So,
    1)-1/3x^3 fluxcion will be -3x^-3=9x^-4 I'm not right?
    What are you doing here? trying to integrate?  \displaystyle \dfrac{-1}{3x^3} \neq -3x^{_3}

    It is written as \displaystyle \dfrac{-1}{3x^3} = \dfrac{-x^{-3}}{3}

    Quote Originally Posted by Kristina View Post
    2)HOwever I have integration simalar like this example and I tried to use it. But I find problem
    Integration (x^2-2) ln(x)dx
    u=x^2-2
    du=2x
    du/dx=2x
    dx=du/2x
    ln(x)/2x*du
    What i do wrong?
    Is your question \displaystyle \int (x^2-2)\;ln(x)\;dx???

    Then start with \displaystyle  \int (x^2-2)\;ln(x)\;dx = \int x^2\;ln(x)\;dx - \int 2\;ln(x)\;dx

    Use integration by parts for the first term!
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  5. #5
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    Okay,
    Now I do not understand nothing...
    integration u*du dx=(x^3/3-2x*1/x (CAN SOMEONE EXPLAIN HOW TO WRITE FORMULAS?)
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  6. #6
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    Quote Originally Posted by Kristina View Post
    Okay,
    Now I do not understand nothing...
    integration u*du dx=(x^3/3-2x*1/x (CAN SOMEONE EXPLAIN HOW TO WRITE FORMULAS?)
    If you posted more clearly and did not confuse your threads with new questions (which I have had to move several times now) you might have a better chance of understanding the help people are giving you.

    Click on the link in my signature for how to format equations.

    Thread closed.
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