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Math Help - Integration help

  1. #1
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    Integration help

    Hi i need to integrate this function:



    I can do the first bit as x^0.5 goes to 2/3x^3/2

    Im really stuck on the second 6/x^3. I need to somehow convert that into a form where i can integrate it. ie, not a fraction. Any suggestions?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nugiboy View Post
    Hi i need to integrate this function:



    I can do the first bit as x^0.5 goes to 2/3x^3/2

    Im really stuck on the second 6/x^3. I need to somehow convert that into a form where i can integrate it. ie, not a fraction. Any suggestions?
    write it as 6x^{-3}

    now what?

    and if your power has more than one character, you must keep them in {} brackets, that is, type x^{0.5} and x^{3/2} as opposed to what you typed. also, you should be aware that LaTeX removes spaces, so that's why what you typed looked so messy. use ~ to insert spaces, or type your text messages in a \text {} or \mbox {}
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  3. #3
    Senior Member Peritus's Avatar
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    (2/3)*x^(3/2) + (-1/2)*6*(x^(-2)) = (2/3)*x^(3/2) -3*(x^(-2))
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    write it as 6x^{-3}

    now what?

    and if your power has more than one character, you must keep them in {} brackets, that is, type x^{0.5} and x^{3/2} as opposed to what you typed. also, you should be aware that LaTeX removes spaces, so that's why what you typed looked so messy. use ~ to insert spaces, or type your text messages in a \text {} or \mbox {}

    Hi thanks for the reply and the help on the Latex.

    Wouldn't 6x^{-3} be 1/{6x^3}? Which is different to 6/{x^3}
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nugiboy View Post
    Hi thanks for the reply and the help on the Latex.

    Wouldn't 6x^{-3} be 1/{6x^3}? Which is different to 6/{x^3}
    no. that would be \frac 16x^{-3} = \frac 1{6x^3}, the -3 power only applies to the x, the 6 has nothing to do with it.
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    no. that would be \frac 16x^{-3} = \frac 1{6x^3}, the -3 power only applies to the x, the 6 has nothing to do with it.
    Ohhhh right. Thanks it all makes sense now!
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