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

1 Answer
Apr 15, 2016

1/((x+7)(x^2+9)) = 1/(58(x+7)) + (7-x)/(58(x^2+9))1(x+7)(x2+9)=158(x+7)+7x58(x2+9)

Explanation:

Working with Real coefficients, the factor x^2+9x2+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))