How do you express $\frac{1}{(x + 7) (x^{2} + 9)}$ $1/((x+7)(x^2+9))$ in partial fractions?

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Answer 1 · 2016-04-15T20:40:32

$\frac{1}{(x + 7) (x^{2} + 9)} = \frac{1}{58 (x + 7)} + \frac{7 - x}{58 (x^{2} + 9)}$ $1/((x+7)(x^2+9)) = 1/(58(x+7)) + (7-x)/(58(x^2+9))$

Working with Real coefficients, the factor $x^{2} + 9$ $x^2+9$ is irreducible over $RR$ since $x^2+9 >= 9$ for all $x in RR$ .

So we want to find a decomposition of the form:

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

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

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

Equating coefficients, we get the following system of simultaneous linear equations:

${(A+B=0),(7B+C=0),(9A+7C=1):}$

Hence:

${(A=1/58),(B=-1/58),(C=7/58):}$

So:

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

How do you express 1/((x+7)(x^2+9))1(x+7)(x2+9)1/((x+7)(x^2+9)) in partial fractions?