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Math Help - Find primitive function

  1. #1
    Senior Member Twig's Avatar
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    Find primitive function

    Hi

    Is this 'supposed' to be solved with partial fractions decomposition?

    Problem:
    Find all primitive functions to:

     \int \frac{1}{x^{2}(x-1)^{3}} \; dx

    Kinda need some starting help if so is the case, thx!
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  2. #2
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    Here is a CAS solution.
    Attached Thumbnails Attached Thumbnails Find primitive function-pf5.gif  
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  3. #3
    Senior Member Twig's Avatar
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    Hi

    im sorry but it is \frac{1}{x^{2}\cdot (x-1)^{3}} and not \frac{1}{x^{3}\cdot (x-1)^{3}}

    But, should I do:

    \frac{1}{x^{2}\cdot (x-1)^{3}} = \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-1)}+\frac{D}{(x-1)^{2}} + \frac{E}{(x-1)^{3}} ?
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  4. #4
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    Quote Originally Posted by Twig View Post
    Hi

    Is this 'supposed' to be solved with partial fractions decomposition?

    Problem:
    Find all primitive functions to:

     \int \frac{1}{x^{2}(x-1)^{3}} \; dx

    Kinda need some starting help if so is the case, thx!
    Try and find constant a - e such that

     <br />
\frac{1}{x^{2}(x-1)^{3}} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x-1} + \frac{d}{(x-1)^2} + \frac{e}{(x-1)^3}<br />

    Yes.
    Last edited by Jester; July 4th 2009 at 01:00 PM. Reason: too slow
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Twig View Post

    Is this 'supposed' to be solved with partial fractions decomposition?
    No.

    Quote Originally Posted by Twig View Post

    Problem:
    Find all primitive functions to:

     \int \frac{1}{x^{2}(x-1)^{3}} \; dx
    Put x=\frac1t and your integral becomes \int{\frac{t^{3}}{(t-1)^{3}}\,dt}=\int{\frac{(t-1)^{3}+(t-1)^{2}+2t(t-1)+(t-1)+1}{(t-1)^{3}}\,dt},

    and I think you can integrate that.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    No.


    Put x=\frac1t and your integral becomes \int{\frac{t^{3}}{(t-1)^{3}}\,dt}=\int{\frac{(t-1)^{3}+(t-1)^{2}+2t(t-1)+(t-1)+1}{(t-1)^{3}}\,dt},

    and I think you can integrate that.
    That's a bit .... final!

    There are several ways of doing it. None are 'right' or 'wrong'. Partial fractions is one way. And there's no doubt about the usefulness and ease of the reciprocal substitution, which is another way.

    However, I suspect that whoever set this question would have had partial fractions in mind.
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