How do you express #(x+4)/(x^2 + 2x + 5)# in partial fractions?

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Answer 1 · 2016-07-12T06:28:24

#(x+4)/(x^2+2x+5) = (1/2-3/4i)/(x+1-2i) + (1/2+3/4i)/(x+1+2i)#

The discriminant #Delta# of the denominator is negative, so it only has Complex zeros:

#Delta = 2^2-4(1)(5) = 4-20 = -16#

The numerator is already linear, so if you want Real coefficients then your expression is already as simple as possible.

On the other hand, you can decompose into simpler fractions using Complex coefficients.

First let's factor the denominator:

#x^2+2x+5#

#=x^2+2x+1+2^2#

#=(x+1)^2-(2i)^2#

#=(x+1-2i)(x+1+2i)#

So we are looking for a solution of the form:

#(x+4)/((x+1-2i)(x+1+2i)) = A/(x+1-2i) + B/(x+1+2i)#

Using Heaviside's cover-up method:

#A = ((-1+2i)+4)/((-1+2i)+1+2i) = (3+2i)/(4i) = 1/2-3/4i#

#B = ((-1-2i)+4)/((-1-2i)+1-2i) = (3-2i)/(-4i) = 1/2+3/4i#

So:

#(x+4)/(x^2+2x+5) = (1/2-3/4i)/(x+1-2i) + (1/2+3/4i)/(x+1+2i)#