# Difficult integral with parameters

• Aug 2nd 2009, 08:12 AM
Dext91
Difficult integral with parameters
Can anyone help me solve this integral?

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

• Aug 2nd 2009, 11:00 AM
DeMath
Quote:

Originally Posted by Dext91
Can anyone help me solve this integral?

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

Hint
$\displaystyle a{x^4} + b{x^3} + c{x^2} + bx + a = 0 \Leftrightarrow a{x^2} + bx + c + \frac{b}{x} + \frac{a}{{{x^2}}} = 0 \Leftrightarrow$

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

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

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

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

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

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

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

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

Next, as is usually, decomposed into two fractions and calculate.
• Aug 2nd 2009, 10:51 PM
centry57
Actually $\displaystyle \sum a_{m}x^{m}=\prod (A_{k}x+B_{k})(C_{n}x^{2}+D_{n}x+K_{n})$
• Aug 3rd 2009, 01:14 AM
DeMath
Quote:

Originally Posted by centry57
Actually $\displaystyle \sum a_{m}x^{m}=\prod (A_{k}x+B_{k})(C_{n}x^{2}+D_{n}x+K_{n})$

(Thinking)

Actually Palindromic polynomial

Do you have a better solution?