How do you write the partial fraction decomposition of the rational expression # 1/((x+6)(x^2+3))#?

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Answer 1 · 2018-03-25T07:53:18

#1/((x+6)(x^2+3)) = 1/(39(x+6))-(x-6)/(39(x^2+3))#

Given:

#1/((x+6)(x^2+3))#

Note that #x^2+3# has no linear factors with real coefficients.

So, assuming we want a decomposition with real coefficients, it takes the form:

#1/((x+6)(x^2+3)) = A/(x+6)+(Bx+C)/(x^2+3)#

Multiplying both sides by #(x+6)(x^2+3)# this becomes:

#1 = A(x^2+3)+(Bx+C)(x+6)#

#color(white)(1) = (A+B)x^2+(6B+C)x+(3A+6C)#

Putting #x=-6# we get:

#1 = A((color(blue)(-6))^2+3) = 39A#

So:

#A=1/39#

Then from the coefficient of #x^2#, we have:

#A+B = 0#

So:

#B = -A = -1/39#

From the constant term we have:

#1 = 3A+6C#

Hence:

#C = 1/6(1-3A) = 1/6(1-1/13) = 2/13 = 6/39#

So:

#1/((x+6)(x^2+3)) = 1/(39(x+6))-(x-6)/(39(x^2+3))#