Difficult integral with parameters

Quote:

Originally Posted by Dext91

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.