$\frac{3 x^2+x+1}{(x-1)^4}=\frac{a}{(x-1)}+\frac{b}{(x-1)^2}$ $+\frac{c}{(x-1)^3}+\frac{d}{(x-1)^4}$
$3 x^2+x+1=a(x-1)^3+b(x-1)^2$ $+c(x-1)+d$
$3 x^2+x+1=a\left(x^3-1+3 x-3 x^2\right)$ $+b\left(x^2+1-2 x\right)+c(x-1)+d$
$3 x^2+x+1=a x^3+(-3 a+b) x^2$ $+(3 a-2 b+c) x+(-a+b-c+d)$
On comparing the coefficient of like powers of $x$ on both sides.
$a=0,-3 a+b=3,3 a-2 b+c=2$
$\Rightarrow \quad b=3 \Rightarrow 0-6+c=1$
$\Rightarrow \quad c=7$
and $\quad-a+b-c+d=1$
$\Rightarrow \quad 0+3-7+d=1$
$\Rightarrow \quad d=5$
Hence, $\quad\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}0 & 3 \\ 7 & 5\end{array}\right]$