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Math Help - Difficult integral with parameters

  1. #1
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    Unhappy Difficult integral with parameters

    Can anyone help me solve this integral?

    Calculate in R
    \int{\frac{dx}{ax^4+bx^3+cx^2+bx+a}}, where b,c Є R and a Є R\{0}.

    Thanks in advance!
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Can anyone help me solve this integral?

    Calculate in R
    \int{\frac{dx}{ax^4+bx^3+cx^2+bx+a}}, where b,c Є R and a Є R\{0}.

    Thanks in advance!
    Hint
    a{x^4} + b{x^3} + c{x^2} + bx + a = 0 \Leftrightarrow a{x^2} + bx + c + \frac{b}{x} + \frac{a}{{{x^2}}} = 0 \Leftrightarrow

    \Leftrightarrow \left( {{x^2} + \frac{1}{{{x^2}}}} \right)a + \left( {x + \frac{1}{x}} \right)b + c = 0.

    {\text{Let }}x + \frac{1}{x} = t \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {t^2} - 2.

    \left( {{t^2} - 2} \right)a + bt + c = 0 \Leftrightarrow a{t^2} + bt - \left( {2a - c} \right) = 0 \Rightarrow

    \Rightarrow {t_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} + 4a\left( {2a - c} \right)} }}{{2a}}.

    x + \frac{1}{x} = {t_{1,2}} \Leftrightarrow {x^2} - {t_{1,2}}x + 1 = 0.

    Then a{x^4} + b{x^3} + c{x^2} + bx + a = a\left( {{x^2} - {t_1}x + 1} \right)\left( {{x^2} - {t_2}x + 1} \right).

    \int {\frac{{dx}}{{a{x^4} + b{x^3} + c{x^2} + bx + a}}}  = \frac{1}{a}\int {\frac{{dx}}{{\left( {{x^2} - {t_1}x + 1} \right)\left( {{x^2} - {t_2}x + 1} \right)}}} .

    If {b^2} + 4a\left( {2a - c} \right) \geqslant 0, then the solution will be on \mathbb{R}.

    Next, as is usually, decomposed into two fractions and calculate.
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  3. #3
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    Actually \sum a_{m}x^{m}=\prod (A_{k}x+B_{k})(C_{n}x^{2}+D_{n}x+K_{n})
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by centry57 View Post
    Actually \sum a_{m}x^{m}=\prod (A_{k}x+B_{k})(C_{n}x^{2}+D_{n}x+K_{n})


    Actually Palindromic polynomial

    Do you have a better solution?
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