Q.Resolve into partial fractions: $\dfrac{x^2+1}{x^2(x-1)}$.

$x^2$ is a repeated linear factor (at $x=0$), so its decomposition needs BOTH a $\dfrac{A}{x}$ term and a $\dfrac{B}{x^2}$ term, alongside the usual term for the simple factor $(x-1)$.

$\dfrac{x^2+1}{x^2(x-1)}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-1}$

$x^2+1=Ax(x-1)+B(x-1)+Cx^2$

At $x=0$: $1=B(-1)\Rightarrow B=-1$

At $x=1$: $2=C(1)\Rightarrow C=2$

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