By observing given equation, the numerator’s degree is lesser compared to the
denominator’s degree.
Let us rewrite the given equation as \(\frac{{{x^2} + 4}}{{x\left( {{x^2} - 4} \right)}} = \frac{{{x^2} + 4}}{{x\left( {x - 2} \right)(x + 2)}}\) --- Equation 1
\(\begin{array}{l}\frac{{{x^2} + 4}}{{x\left( {x - 2} \right)(x + 2)}} = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}}\\\\\frac{{{x^2} + 4}}{{x\left( {x - 2} \right)(x + 2)}} = \frac{{A(x + 2)(x - 2) + B(x)(x - 2) + C(x)(x + 2)}}{{x(x - 2)(x + 2)}}\\\end{array}\)
\({x^2} + 4 = A({x^2} - 4) + B({x^2} - 2x) + C({x^2} + 2x)\) --- Equation 2
From Equation 2 Let us equate both the sides based on the degrees.
Considering the degree 2 we get the equation, \(1 = A + B + C\) -- Equation 3
Considering the degree 1 we get the equation, \(0 = - 2B + 2C\) -- Equation 4
From Equation 4 we can conclude that \(B = C\) -- Equation 5
Considering the degree 1 we get the equation, \(4 = - 4A\) -- Equation 6
From Equation 5 we can conclude that the value of \(A = - 1\)
By applying the value of \(A = - 1\) in Equation 3 we get
\(1 = - 1 + B + C\) -- Equation 7
We know that the value of \(B = C\) so the Equation 7 will become \(2 = 2B\)
Hence the value of \(B = 1\) and \(C = 1\)
We can rewrite the Equation 1 by applying the value of \(A,B,C\) as
\(\frac{{{x^2} + 4}}{{x\left( {{x^2} - 4} \right)}} = \frac{{ - 1}}{x} + \frac{1}{{x + 2}} + \frac{1}{{x - 2}}\)
Thus, we can write the Equation\(\frac{{{x^2} + 4}}{{x\left( {{x^2} - 4} \right)}}\) in the form of \(\frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}}\)
