Respuesta :

LammettHash
The denominator is equivalent to

[tex](x^2+4)^3(x+3)^2(x-3)^2(x-1)(x^2+x+1)[/tex]

The decomposition would take the form

[tex]\dfrac{a_1x+a_2}{x^2+4}+\dfrac{a_3x+a_4}{(x^2+4)^2}+\dfrac{a_5x+a_6}{(x^2+4)^3}+\dfrac{a_7}{x+3}+\dfrac{a_8}{(x+3)^2}+\dfrac{a_9}{x-3}+\dfrac{a_{10}}{(x-3)^2}+\dfrac{a_{11}}{x-1}+\dfrac{a_{12}x+a_{13}}{x^2+x+1}[/tex]

Otras preguntas