Answer:
[tex]\dfrac{2x+4}{x(x+2)(x-4)} \equiv \dfrac{1}{2(x-4)}-\dfrac{1}{2x}[/tex]
Step-by-step explanation:
Partial Fractions
Write out the expression as an identity:
[tex]\begin{aligned}\dfrac{2x+4}{x(x+2)(x-4)} & \equiv \dfrac{A}{x}+\dfrac{B}{(x+2)}+\dfrac{C}{(x-4)}\\\\\implies \dfrac{x(2x+4)(x+2)(x-4)}{x(x+2)(x-4)} & \equiv \dfrac{Ax(x+2)(x-4)}{x}+\dfrac{Bx(x+2)(x-4)}{(x+2)}+\dfrac{Cx(x+2)(x-4)}{(x-4)}\\\\\implies 2x+4 & \equiv A(x+2)(x+4)+ Bx(x-4)+Cx(x+2)\end{aligned}[/tex]
Calculate the values of A, B and C using substitution:
[tex]\begin{aligned}2x+4 & = A(x+2)(x-4)+Bx(x-4)+Cx(x+2)\\\\x=4 \implies 12 & = A(0)+B(0)+C(24)\implies C=\dfrac{1}{2}\\\\x=-2 \implies 0 & = A(0)+B(12)+C(0) \implies B=0\\\\ x=0 \implies 4 & = A(-8)+B(0)+C(0) \implies A=-\dfrac{1}{2}\end{aligned}[/tex]
Replace A, B and C in the original identity:
[tex]\begin{aligned}\dfrac{2x+4}{x(x+2)(x-4)} & \equiv \dfrac{A}{x}+ \dfrac{B}{(x+2)}+\dfrac{C}{(x-4)}\\\\& \equiv -\dfrac{1}{2x}+\dfrac{1}{2(x-4)}\\\\\implies \dfrac{2x+4}{x(x+2)(x-4)}& \equiv \dfrac{1}{2(x-4)}-\dfrac{1}{2x}\end{aligned}[/tex]
