Answer:
a.[tex]\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+5}[/tex]
b.[tex]\frac{A}{x+3}+\frac{B}{x-3}+\frac{C}{(x-3)^2}+\frac{D}{(x+3)^2}[/tex]
Explanation:
a.We are given that
[tex]\frac{x^4+7}{x^5+5x^3}[/tex]
[tex]\frac{x^4+7}{x^5+5x^3}=\frac{x^4+7}{x^3(x^2+5)}[/tex]
Using partial fraction decomposition of the given function
[tex]\frac{x^4+7}{x^3(x^2+5)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+5}[/tex]
Using the formula
[tex]\frac{1}{x^3(x^2+a)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+a}[/tex]
b.[tex]\frac{2}{(x^2-9)^2}[/tex]
[tex]\frac{2}{(x^2-3^2)^2}=\frac{2}{(x+3)^2(x-3)^2}[/tex]
Using property [tex]a^2-b^2=(a+b)(a-b)[/tex]
[tex]\frac{2}{(x+3)^2(x-3)^2}=\frac{A}{x+3}+\frac{B}{x-3}+\frac{C}{(x-3)^2}+\frac{D}{(x+3)^2}[/tex]
Using the property
[tex]\frac{1}{x^2}=\frac{A}{x}+\frac{B}{x^2}[/tex]