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    \int 1-x^3 \div x^2 \ dx

    I thought that I would be best served by splitting the fraction into two. Therefore, I had 1/x^2; which I identified as being equal to x^-2. Adding one to that; simplified that, and I got -1/x

    The second part of my "solution" was -x^3/x^2. I thought that because there was indices and division involved, this meant that I would subtract the indices from one another, leaving me with -x^1. Adding one to that gave me -x^2.

    However; the online integral calculator has listed the result as -x^2/2
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    The x terms can be simplified.

    The 1 obviously turns into an x. You should be sorted.

    If all divided by x^2 then.... Integral of 1/x^2 - x after you've simplified the x terms. Then 1/x^2 = x^-2.... so Integral of x^-2 - x then integrate and add +c
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    (Original post by apronedsamurai)
    \int 1-x^3 \div x^2 \ dx
    Is all of the 1-x^3 divided by the x^2

    If so

    (1-x^3)/x^2 = 1/x^2 - x

    1/x^2 is the same as x^(-2)

    So

    Once again

    Increase the power and divide
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    What IS the proper form for -x^2?

    I had thought it was - 1/x^2
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    (Original post by apronedsamurai)
    What IS the proper form for -x^2?

    I had thought it was - 1/x^2
    Is this a different question
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    No,

    I am trying to narrow down where I have went wrong.
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    (Original post by apronedsamurai)
    No,

    I am trying to narrow down where I have went wrong.
    Well this question does not include -x^2

    1/(x^2) = x^(-2)
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    (Original post by apronedsamurai)
    \int 1-x^3 \div x^2 \ dx

    I thought that I would be best served by splitting the fraction into two. Therefore, I had 1/x^2; which I identified as being equal to x^-2. Adding one to that; simplified that, and I got -1/x

    The second part of my "solution" was -x^3/x^2. I thought that because there was indices and division involved, this meant that I would subtract the indices from one another, leaving me with -x^1. Adding one to that gave me -x^2.

    However; the online integral calculator has listed the result as -x^2/2
    The integral of -x is indeed -x^2/2

    You seem to have forgotten the basic rule about dividing by the new power when you integrate something of the form x^n - you get \dfrac{x^{n+1}}{n+1}

    The way to check these things is ALWAYS to differentiate your final answer and check that you get back the original function.
 
 
 
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