Question

Write the partial fraction decomposition fro the ration expression below

Answer 1

$\dfrac{5x-2}{(x-1)^2}=\dfrac a{x-1}+\dfrac b{(x-1)^2}$
$5x-2=a(x-1)+b$

Since $5x-2=5x-5+3=5(x-1)+3$, it follows that $a=5$ and $b=3$, so

$\dfrac{5x-2}{(x-1)^2}=\dfrac5{x-1}+\dfrac3{(x-1)^2}$

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac{x^3+x+x+2}{x^2(x^2+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac{cx+d}{x^2+1}$
$x^3+x^2+x+2=ax(x^2+1)+b(x^2+1)+(cx+d)x^2$
$x^3+x^2+x+2=(a+c)x^3+(b+d)x^2+ax+b$

When $x=0$, you find that $b=2$. It's also clear that $a=1$. So the remaining constants must be $1+c=1\implies c=0$ and $2+d=1\implies d=-1$, and so you get

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac1x+\dfrac2{x^2}-\dfrac1{x^2+1}$