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∫(x-1∕x)² dx Help, please Watch

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    The question is in the title.
    So far I have written it as
    ∫x²-(x-1)²dx
    and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
    I've done the first x
    ∫x² dx =x³/3
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    (Original post by TheGreatPumpkin)
    The question is in the title.
    So far I have written it as
    ∫x²-(x-1)²dx
    and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
    I've done the first x
    ∫x² dx =x³/3
    Expand (x- \frac{1}{x})^2 = (x-\frac{1}{x})(x-\frac{1}{x})

    The result of (a+b)^2 is a^2 +2ab +b^2, that is the first term squared, plus two times the product, plus the last term squared.

     \int (x- \frac{1}{x})^2 dx = \int(x^2 - 2 + \frac{1}{x^2})dx

    From here it can be integrated easily.
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    (Original post by TheGreatPumpkin)
    The question is in the title.
    So far I have written it as
    ∫x²-(x-1)²dx
    and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
    I've done the first x
    ∫x² dx =x³/3
    You can't go from ∫(x-1∕x)² to ∫x^2 - (x^-1)^2. You need to use the reverse chain rule so you first need to differentiate the bracket so you get dy/dx = 1 + x^-2. Using the reverse chain rule you get ((x-1/x)^3)/(3(1+x^-2)). Any math geniuses agree? Quite new to integration so sorry for any mistake. Hope it helps
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    (Original post by NotNotBatman)
    Expand (x- \frac{1}{x})^2 = (x-\frac{1}{x})(x-\frac{1}{x})

    The result of (a+b)^2 is a^2 +2ab +b^2, that is the first term squared, plus two times the product, plus the last term squared.

     \int (x- \frac{1}{x})^2 dx = \int(x^2 - 2 + \frac{1}{x^2})dx

    From here it can be integrated easily.
    Any help in what I did wrong?
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    (Original post by samendrag)
    Any help in what I did wrong?
    Yeah, (x-\frac{1}{x})^2 \not= x^2-(x^{-1})^2
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    (Original post by RDKGames)
    Yeah, (x-\frac{1}{x})^2 \not= x^2-(x^{-1})^2
    Could you use the reverse chain rule, how would you use it in this question?
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    (Original post by samendrag)
    You can't go from ∫(x-1∕x)² to ∫x^2 - (x^-1)^2. You need to use the reverse chain rule so you first need to differentiate the bracket so you get dy/dx = 1 + x^-2. Using the reverse chain rule you get ((x-1/x)^3)/(3(1+x^-2)). Any math geniuses agree? Quite new to integration so sorry for any mistake. Hope it helps
    (Original post by samendrag)
    Any help in what I did wrong?
    x is a variable factor, so the reverse chain rule doesn't apply. You can divide by constants when using the reverse chain rule, but not by variable factors.
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    (Original post by NotNotBatman)
    x is a variable factor, so the reverse chain rule doesn't apply. You can divide by constants when using the reverse chain rule, but not by variable factors.
    ohh ok thanks!
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    (Original post by NotNotBatman)
    Expand (x- \frac{1}{x})^2 = (x-\frac{1}{x})(x-\frac{1}{x})

    The result of (a+b)^2 is a^2 +2ab +b^2, that is the first term squared, plus two times the product, plus the last term squared.

     \int (x- \frac{1}{x})^2 dx = \int(x^2 - 2 + \frac{1}{x^2})dx

    From here it can be integrated easily.
    Thanks, I forgot how to expand for a moment
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    (Original post by samendrag)
    Could you use the reverse chain rule, how would you use it in this question?
    Apologies, that reply was meant for OP. The chain rule wouldn't be of much help here.
 
 
 
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