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Math Help - Integration Please help,

  1. #1
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    Integration Please help,

    Hello Please help me,
    Can someone tell me if I have gone wrong in my attached drawing,I have to do integration by partial fractions,also any tips on methods to do partial fraction method would be great, I am not very good at them
    Attached Thumbnails Attached Thumbnails Integration Please help,-pfint.jpg  
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  2. #2
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    Quote Originally Posted by wolfhound View Post
    Hello Please help me,
    Can someone tell me if I have gone wrong in my attached drawing,I have to do integration by partial fractions,also any tips on methods to do partial fraction method would be great, I am not very good at them
    Your values for A & B are correct.
    But you did not substitute this values in the original equation correctly.
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  3. #3
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    Hello miss,
    Can you tell me how please?
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  4. #4
    Member Miss's Avatar
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    Quote Originally Posted by wolfhound View Post
    Hello miss,
    Can you tell me how please?
    You have the equation:
    \frac{1}{x^2-2x-1}=\frac{A}{x-1+\sqrt{2}}+\frac{B}{x-1-\sqrt{2}}

    And A=\frac{1}{-2\sqrt{2}} \,\ B=\frac{1}{2\sqrt{2}}


    So when you substitute these values of A&B in the equation, what will you get?
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  5. #5
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    Red face

    Quote Originally Posted by Miss View Post
    You have the equation:
    \frac{1}{x^2-2x-1}=\frac{A}{x-1+\sqrt{2}}+\frac{B}{x-1-\sqrt{2}}

    And A=\frac{1}{-2\sqrt{2}} \,\ B=\frac{1}{2\sqrt{2}}


    So when you substitute these values of A&B in the equation, what will you get?
    I dont know what way to arrange them on top of the fractions?
    Help please! Thanks
    Is it 1/(x-1-Sqrt2) + 1/(x-1+Sqrt2)
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  6. #6
    Member Miss's Avatar
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    Quote Originally Posted by wolfhound View Post
    I dont know what way to arrange them on top of the fractions?
    Help please! Thanks
    Is it 1/(x-1-Sqrt2) + 1/(x-1+Sqrt2)
    Don't make it complicated
    Just replace A by its value and B by its value.
    It will be :
    \frac{ \frac{1}{-2\sqrt{2}} }{ x-1+\sqrt{2} } + \frac{ \frac{1}{2\sqrt{2}} }{ x - 1 - \sqrt{2} }.

    To make our life easier, we can write it as:

    \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}}

    So:

    \int \frac{1}{x^2-2x-1} \, dx = \int \left( \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}} \right) \, dx.

    Can you integrate this?
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  7. #7
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    Smile

    Quote Originally Posted by Miss View Post
    Don't make it complicated
    Just replace A by its value and B by its value.
    It will be :
    \frac{ \frac{1}{-2\sqrt{2}} }{ x-1+\sqrt{2} } + \frac{ \frac{1}{2\sqrt{2}} }{ x - 1 - \sqrt{2} }.

    To make our life easier, we can write it as:

    \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}}

    So:

    \int \frac{1}{x^2-2x-1} \, dx = \int \left( \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}} \right) \, dx.

    Can you integrate this?
    Hello miss
    Thanks, I can look up how to integrate this,but you could show me if you like
    Thanks for your help!
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  8. #8
    Member Miss's Avatar
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    Quote Originally Posted by wolfhound View Post
    Hello miss
    Thanks, I can look up how to integrate this,but you could show me if you like
    Thanks for your help!
    \int \left( \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}} \right) \, dx

    =\frac{-1}{2\sqrt{2}} \int \frac{dx}{x-1+\sqrt{2}} \, + \frac{1}{2\sqrt{2}} \int \frac{dx}{x-1-\sqrt{2}} \,

    Can you complete it ?
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  9. #9
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    Quote Originally Posted by Miss View Post
    \int \left( \frac{-1}{2\sqrt{2}} \, \frac{1}{x-1+\sqrt{2}} + \frac{1}{2\sqrt{2}} \, \frac{1}{x-1-\sqrt{2}} \right) \, dx

    =\frac{-1}{2\sqrt{2}} \int \frac{dx}{x-1+\sqrt{2}} \, + \frac{1}{2\sqrt{2}} \int \frac{dx}{x-1-\sqrt{2}} \,

    Can you complete it ?
    Yes I have it
    Thanks
    Last edited by wolfhound; March 4th 2010 at 01:58 PM.
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  10. #10
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    You could compare the answer you get to

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

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

    which is of the form

    \int{\frac{1}{a^2-u^2}}du=\frac{1}{2a}ln|\frac{a+u}{a-u}|

    =-\frac{1}{2\sqrt{2}}ln|\frac{\sqrt{2}+x-1}{\sqrt{2}-x+1}|
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