Answer:
Step-by-step explanation:
Given is a rational algebraic expression in x as
[tex]\frac{5x+2}{x^4-4x^2}[/tex]
First let us factorize the denominator.
[tex]x^2(x^2-4)\\=x^2(x+2)(x-2)[/tex]
Let the given term be
[tex]\frac{A}{x} +\frac{B}{x^2} +\frac{C}{x-2} +\frac{c}{x+2} \\\\=\frac{Ax(x^2-4)+B(x^2-4)+cx^2(x+2)+Dx^2(x-2)}{x^4-4x^2}[/tex]
Equate the numerator to get
[tex]Ax(x^2-4)+B(x^2-4)+cx^2(x+2)+Dx^2(x-2)[/tex]=[tex]5x+2[/tex]
Substitute x =-2
[tex]4D (-4) = -8\\D=\frac{1}{2}[/tex]
Now substitute x =2
We get
[tex]16C=12\\C=\frac{3}{4}[/tex]
Put x=0
[tex]-4B=2\\B=\frac{-1}{2}[/tex]
Next equate X cube terms to 0 on right side
[tex]A+C+D=0\\A=\frac{-5}{4}[/tex]
Hence partial fraction is
[tex]\frac{-1}{2x^2} -\frac{5}{4x} +\frac{1}{2(x+2)} +\frac{3}{4(x-2)}[/tex]
