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Thread: Integration using partial fractions.

  1. #1
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    Integration using partial fractions.

    could someone help me

    (x+5)dx/(x^2+x+1)

    and this

    (1/sqrt(2)*x+1)dx/(x^2-x*sqrt(2) +1)
    Last edited by jo-ja; Nov 2nd 2009 at 02:37 AM.
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  2. #2
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    Quote Originally Posted by jo-ja View Post
    could someone help me

    (x+5)dx/(x^2+x+1)
    By completing the square: $\displaystyle x^2+x+1= x^2+x+ (1/4)- (1/4)+ 1= (x+1/2)^2+ 3/4$. That cannot be factored in terms of real numbers and so you cannot use "partial fractions". Make the substitution u= x+ 1/2 instead.

    and this

    (1/sqrt(2)*x+1)dx/(x^2-x*sqrt(2) +1)
    Here, completing the square gives $\displaystyle x^2- x\sqrt{2}+ 1/2- 1/2+ 1= (x-\frac{\sqrt{2}}{2})^2+ 3/4$.

    Again, you can't use partial fractions. Make the substitution $\displaystyle u= x-\frac{\sqrt{2}}{2}$ instead.
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  3. #3
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    My bad

    originaly was dx/(x^4+1) so then =>
    dx/(x^2+x*sqrt(2)+1)*(x^2-x*sqrt(2)+1)
    but then ....
    Plz help
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  4. #4
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    Consider

    $\displaystyle \int \frac{x^2 + 1 }{x^4 + 1}~dx$

    $\displaystyle = \int \frac{ 1 + 1/x^2 }{ x^2 + 1/x^2}~dx $

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

    Sub $\displaystyle x - 1/x = t ~~\implies (1 + 1/x^2 )dx = dt$

    the integral becomes

    $\displaystyle \int \frac{dt}{ t^2 + 2}$

    $\displaystyle = \frac{1}{\sqrt{2}} \tan^{-1}( \frac{x - 1/x}{\sqrt{2}}) + c$

    similarly for $\displaystyle \int \frac{ x^2 - 1}{x^4 + 1}~dx $

    $\displaystyle = - \frac{1}{\sqrt{2}} \tanh^{-1}( \frac{\sqrt{2}}{x + 1/x}) + c$

    Finally , take sum of difference of them , what will you see then ?
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  5. #5
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    I dont understand nothing coud someone plz solve mi this integral to the end
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  6. #6
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    Quote Originally Posted by jo-ja View Post
    I dont understand nothing coud someone plz solve mi this integral to the end

    Here is the technique of resolving $\displaystyle \frac{1}{1 + x^n} $ into partial fractions .

    $\displaystyle (1 + x^n ) = (1 +x ) ( 1 - ax + x^2 ) ( 1 - bx + x^2 ) ... $

    ( * the term " 1 + x " exists only for n is odd )
    Then , " log " both sides

    $\displaystyle \ln(1 + x^n) = \ln(x+1) + \ln(1 - ax + x^2) + \ln(1 - bx + x^2 ) + ...$

    After that , differentiate both sides ( we call this " Logarithmic Differentiaion ! )

    $\displaystyle \frac{ nx^{n-1}}{ 1 + x^n} = \frac{1}{1 + x} + \frac{ 2x - a}{1 - ax + x^2} + \frac{2x - b}{ 1 - bx + x^2} + ... $


    Replace $\displaystyle x $ with $\displaystyle \frac{1}{x}$

    $\displaystyle \frac{ n x }{ 1 + x^n } = \frac{x}{1 + x} + \frac{ 2x - ax^2 }{ 1 - ax + x^2 } + \frac{ 2x - bx^2 }{ 1 - bx + x^2 } + ...$

    Divide both sides by $\displaystyle x $

    $\displaystyle \frac{1}{ 1 + x^n } = \frac{1}{n} [ \frac{1}{ 1 + x } + \frac{ 2 - ax }{1 -ax + x^2} + \frac{ 2 -bx}{ 1 -bx + x^2} + ...$

    ps : you should know what a ,b , c , d ... are . for this case ,
    let $\displaystyle a = \sqrt{2} , b = -\sqrt{2}$
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  7. #7
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    problem solved, tnx evryone, you are the best!!!!!
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