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Math Help - Integration by partial fractions

  1. #1
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    Integration by partial fractions

    Hi all!
    I am working on this Integration by partial fraction question and I used the -b formula to factor the bottom line but I'm not sure where to go from here. I would really appreciate any help!

    \int \frac{1}{x^2-2x-1}dx
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  2. #2
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    Quote Originally Posted by NFZ17 View Post
    Hi all!
    I am working on this Integration by partial fraction question and I used the -b formula to factor the bottom line but I'm not sure where to go from here. I would really appreciate any help!

    \int \frac{1}{x^2-2x-1}dx
    Hi NFZ17,

    so you have x^2-2x-1=(x-1-\sqrt{2})(x-1+\sqrt{2})

    then

    \frac{1}{x^2-2x-1}=\frac{A}{x-1-\sqrt{2}}+\frac{B}{x-1+\sqrt{2}}

    1=A(x-1+\sqrt{2})+B(x-1-\sqrt{2})

    Finish up by placing x=1+\sqrt{2} to find A, and x=1-\sqrt{2} to find B
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  3. #3
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    Quote Originally Posted by NFZ17 View Post
    Hi all!
    I am working on this Integration by partial fraction question and I used the -b formula to factor the bottom line but I'm not sure where to go from here. I would really appreciate any help!

    \int \frac{1}{x^2-2x-1}dx
    Alternatively:

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

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


    Now you can make the substitution x - 1 = \sqrt{2}\cosh{t} so that dx = \sqrt{2}\sinh{t}\,dt.

    Note that t = \textrm{arcosh}\,{\left(\frac{\sqrt{2}(x - 1)}{2}\right)}.


    The integral becomes:

    \int{\frac{1}{(\sqrt{2}\cosh{t})^2 - 2}\,\sqrt{2}\sinh{t}\,dt}

     = \int{\frac{\sqrt{2}\sinh{t}}{2\cosh^2{t} - 2}\,dt}

     = \int{\frac{\sqrt{2}\sinh{t}}{2(\cosh^2{t} - 1)}\,dt}

     = \int{\frac{\sqrt{2}\sinh{t}}{2\sinh^2{t}}\,dt}

     = \int{\frac{\sqrt{2}}{2}\,dt}

     = \frac{\sqrt{2}}{2}t + C

     = \frac{\sqrt{2}}{2}\textrm{arcosh}\,{\left(\frac{\s  qrt{2}(x - 1)}{2}\right)} + C.
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    Thanks a million lads!
    Archie, I used your method and came up with...

    \frac{1}{2\sqrt2} ln[x-1+\sqrt2]-\frac{1}{2\sqrt2} ln[x-1-\sqrt2]+c

    Does this look right? I'm a disaster when it comes to surds!
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  5. #5
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    Hi NFZ17,

    double check your values for A and B to see if you've got the two
    parts with surds the right way around in your final answer.
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    Hi NFZ17,

    so you have x^2-2x-1=(x-1-\sqrt{2})(x-1+\sqrt{2})

    then

    \frac{1}{x^2-2x-1}=\frac{A}{x-1-\sqrt{2}}+\frac{B}{x-1+\sqrt{2}}

    1=A(x-1+\sqrt{2})+B(x-1-\sqrt{2})

    Finish up by placing x=1+\sqrt{2} to find A, and x=1-\sqrt{2} to find B
    If I set x=1+\sqrt{2}

    I get, A=0 and, B=\frac{1}{2\sqrt2}

    and then,

    A=\frac{-1}{2\sqrt2}

    is this where I'm going wrong?
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  7. #7
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    Quote Originally Posted by NFZ17 View Post
    If I set x=1+\sqrt{2}

    I get, A=0 and, B=\frac{1}{2\sqrt2}

    and then,

    A=\frac{-1}{2\sqrt2}

    is this where I'm going wrong?
    If you set x=1+\sqrt{2}, you get

    1=A(1+\sqrt{2}-1+\sqrt{2})+B(1+\sqrt{2}-1-\sqrt{2})=A(2\sqrt{2})

    Then A=\frac{1}{2\sqrt{2}}

    You get A through multiplying B by 0.
    Similarly, find B by cancelling A.
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  8. #8
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    Thank Archie!

    \frac{1}{2\sqrt2} ln[x-1-\sqrt2]-\frac{1}{2\sqrt2} ln[x-1+\sqrt2]+c

    I hope this is right!
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