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

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Answer 1 · 2016-04-15T20:40:32

#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# 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))#