Question

partial fraction decomposition:
(2x^3+4x-1)/(x^2+2)^2

Answer 1

\[\frac{2x^3+4x-1}{(x^2+2)^2}=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{(x^2+2)^2}\]
or you can go the longer way using
\[\frac{2x^3+4x-1}{(x^2+2)^2}=\frac{Ax+B}{x^2+2}+\frac{Cx^3+Dx^2+Ex+F}{(x^2+2)^2}\]
I'll refer to the first way.
\[\begin{align*}\frac{2x^3+4x-1}{(x^2+2)^2}&=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{(x^2+2)^2}\\\\
2x^3+4x-1&=(Ax+B)(x^2+2)+(Cx+D)\\\\
2x^3+4x-1&=Ax^3+Bx^2+2Ax+2B+Cx+D\\\\
2x^3+4x-1&=Ax^3+Bx^2+(2A+C)x+(2B+D)
\end{align*}\]
yielding the system
\[\begin{cases}
A=2\\B=0\\2A+C=4\\2B+D=-1
\end{cases}~~\Rightarrow~~A=2,~B=0,~C=0,~D=-1\]