Answer
a. It is already a partial fraction decomposition
b. $\frac{x}{(x+1)^2}=\frac{1}{x+1}+\frac{-1}{(x+1)^2}$
c. It is already a partial fraction decomposition
d. It is already a partial fraction decomposition
Work Step by Step
a. It is already a partial fraction decomposition
b. It can be decomposed further.
$\frac{x}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}$,
$A(x+1)+B$,
$Ax+A+B=x$,
$
A=1,\\
1+B=0, B=-1
$.
Therefore,
$\frac{x}{(x+1)^2}=\frac{1}{x+1}+\frac{-1}{(x+1)^2}$
c. It is already a partial fraction decomposition
d. $\frac{x+2}{(x^2+1)^2}=\frac{Ax+b}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$,
$A(x^2+1)+Cx+D$,
$Ax^2+A+Cx+D$,
$Ax^2+Cx+A+D=x+2$,
thus, $A=0, B=0, A+C=1, C=1, B+D=2, D=2$.
Therefore,
$\frac{x+2}{(x^2+1)^2}=\frac{x+2}{(x^2+1)^2}$,
Therefore, it is already a partial fraction decomposition.
