Results 1 to 8 of 8

Thread: Integration by partial fractions

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    13

    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!

    $\displaystyle \int \frac{1}{x^2-2x-1}dx$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    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!

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

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

    then

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

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

    Finish up by placing $\displaystyle x=1+\sqrt{2}$ to find A, and $\displaystyle x=1-\sqrt{2}$ to find B
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    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!

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

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

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


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

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


    The integral becomes:

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

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

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

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

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

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

    $\displaystyle = \frac{\sqrt{2}}{2}\textrm{arcosh}\,{\left(\frac{\s qrt{2}(x - 1)}{2}\right)} + C$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2010
    Posts
    13
    Thanks a million lads!
    Archie, I used your method and came up with...

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

    Does this look right? I'm a disaster when it comes to surds!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2010
    Posts
    13
    Quote Originally Posted by Archie Meade View Post
    Hi NFZ17,

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

    then

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

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

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

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

    and then,

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

    is this where I'm going wrong?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by NFZ17 View Post
    If I set $\displaystyle x=1+\sqrt{2}$

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

    and then,

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

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

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

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

    You get A through multiplying B by 0.
    Similarly, find B by cancelling A.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2010
    Posts
    13
    Thank Archie!

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

    I hope this is right!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial Fractions/ Integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 13th 2010, 03:53 PM
  2. Integration by partial fractions?
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Oct 22nd 2009, 09:03 AM
  3. integration using partial fractions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 4th 2009, 02:26 AM
  4. Integration using partial fractions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 16th 2009, 10:34 AM
  5. Integration by partial fractions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 8th 2007, 01:42 PM

Search Tags


/mathhelpforum @mathhelpforum