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    i have to integrate x^-4 with res pect to x.
    so i got (x/-3)^-3. How do i simplify this answer that has x over -3? as in make the answer have no fractions?
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    (Original post by Chelsea12345)
    i have to integrate x^-4 with res pect to x.
    so i got (x/-3)^-3. How do i simplify this answer that has x over -3? as in make the answer have no fractions?
    You should have obtained \displaystyle \frac{x^{-3}}{-3} + k which is significantly different from the answer you posted.

    This can be rewritten as \displaystyle k-\frac{1}{3x^3} if that helps at all.
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    (Original post by Chelsea12345)
    i have to integrate x^-4 with res pect to x.
    so i got (x/-3)^-3. How do i simplify this answer that has x over -3? as in make the answer have no fractions?
    But \displaystyle \int x^{-n} .dx= \frac{x^{-n+1}}{-n+1} +c \not= (\frac{x}{-n+1})^{-n+1} for n\not= 1
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    (Original post by RDKGames)
    ...
    Quick! Fix that. I'm getting the shivers.

    Edit: dx and logarithmic shivers are passing but I still have an arbitrary constant twitch.
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    most books put +c rather than +k in case you were wondering

    :dontknow:
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    (Original post by Mr M)
    Quick! Fix that. I'm getting the shivers.
    Being stuck on a question for the last 3 hours seems to have done my head in :lol:
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    (Original post by RDKGames)
    But \displaystyle \int x^{-n} .dx= \frac{x^{-n+1}}{-n+1} +c \not= (\frac{x}{-n+1})^{-n+1} for n\not= 1
    How are you guys able to write in fractions on here?
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    (Original post by the bear)
    most books put +c rather than +k in case you were wondering

    :dontknow:
    The authors of those books must be pretty unkool.
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    (Original post by Chelsea12345)
    How are you guys able to write in fractions on here?
    https://www.thestudentroom.co.uk/wiki/LaTex
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    (Original post by Mr M)
    The authors of those books must be pretty unkool.
    they are not down with the cids

    :five:
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    thankyouuu
 
 
 
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