How do you write the partial fraction decomposition of the rational expression $1/((x+7)(x^2+9))$ ?

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Answer 1 · 2015-12-14T03:40:38

$1/((x+7)(x^2+9))=(7-x)/(58(x^2+9))+1/(58(x+7))$

$1/((x+7)(x^2+9))=A/(x+7)+(Bx+C)/(x^2+9)$

$1=A(x^2+9)+(Bx+C)(x+7)$

IF $x=-7$ :

$1=58A$
$A=1/58$

IF $x=0$ :

$1=9/58+7C$
$C=7/58$

IF $x=7$ :

$1=1+14(7B+7/58)$
$B=-1/58$

Plug in these values:

$1/((x+7)(x^2+9))=(7-x)/(58(x^2+9))+1/(58(x+7))$

How do you write the partial fraction decomposition of the rational expression 1/((x+7)(x^2+9))1/((x+7)(x^2+9))?