Answer
$a.\displaystyle \qquad x^{4}+4x^{2}+16+\frac{A}{x+2}+\frac{B}{x-2}$
$b.\qquad \displaystyle \frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^{2}+1}+\frac{Ex+F}{(x^{2}+1)^{2}}$
Work Step by Step
$a.$
The denominator is a difference of squares , $(x-2)(x+2).\\\\$
The numerator has degree grreater than the denominator. Long division:
$ \begin{array}{llllll}
& x^{4} & +4x^{2} & +16 & & \\
& -- & -- & -- & -- & \\
(x^{2}-4) & )x^{6} & & & & \\
& x^{6} & -4x^{4} & & & \\
& -- & -- & -- & & \\
& & 4x^{4} & & & \\
& & 4x^{4} & -16x^{2} & & \\
& & -- & -- & & \\
& & & 16x^{2} & & \\
& & & 16x^{2} & +64 &
\end{array}$
$\displaystyle \frac{x^{6}}{x^{2}-4}=x^{4}+4x^{2}+16+\frac{64}{(x+2)(x-2)}$
$= x^{4}+4x^{2}+16+\displaystyle \frac{A}{x+2}+\frac{B}{x-2}$
$b.$
Denominator: $x(x-1)(x^{2}+1)^{2}$ has degree $6$, numerator has $5$.
There are repeated factors in the denominator.
$\displaystyle \frac{x^{5}+1}{x(x-1)(x^{2}+1)^{2}}= \displaystyle \frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^{2}+1}+\frac{Ex+F}{(x^{2}+1)^{2}}$
