How do you express (x+4)/[(x+1)^2 + 4] in partial fractions?
1 Answer
With Real coefficients:
(x+4)/((x+1)^2+4) = (x+4)/(x^2+2x+5)
With Complex coefficients:
(x+4)/((x+1)^2+4) = (2-3i)/(4(x+1-2i))+(2+3i)/(4(x+1+2i))
Explanation:
Note that if
Hence:
So if we are restricted to Real coefficients then this does not break down into simpler partial fractions.
We can expand and render the denominator in standard form:
(x+4)/((x+1)^2+4) = (x+4)/(x^2+2x+5)
On the other hand, if we want to use Complex coefficients then the denominator factors as a difference of squares:
(x+1)^2+4 = (x+1)^2-(2i)^2 = (x+1-2i)(x+1+2i)
So for some
(x+4)/((x+1)^2+4) = A/(x+1-2i)+B/(x+1+2i)
=(A(x+1+2i)+B(x+1-2i))/((x+1)^2+4)
=((A+B)x+(A(1+2i)+B(1-2i)))/((x+1)^2+4)
Equating coefficients:
{ (A+B=1), (A(1+2i)+B(1-2i) = 4) :}
Subtracting the first equation from the second we get:
2i(A-B) = 3
So:
A-B = 3/(2i) = -3/2i
Hence:
A = 1/2-3/4i
B = 1/2+3/4i
So:
(x+4)/((x+1)^2+4) = (2-3i)/(4(x+1-2i))+(2+3i)/(4(x+1+2i))