$\frac{3x^2+x+1}{{(x-1)}^4}=\frac{a}{(x-1)}+\frac{b}{{(x-1)}^2}+\frac{c}{{(x-1)}^3}+\frac{d}{{(x-1)}^4}$ are

$\frac{3x^2+x+1}{{(x-1)}^4}=\frac{a}{(x-1)}+\frac{b}{{(x-1)}^2}+\frac{c}{{(x-1)}^3}+\frac{d}{{(x-1)}^4}$

$x-1=y\Rightarrow x=y+1$

$\frac{3x^2+x+1}{{(x-1)}^4}=\frac{3{(y+1)}^2+y+1+1}{{\left(y\right)}^4}$

$=\frac{3(y^2+2y+1)+y+2}{y^4}$

$=\frac{3y^2+6y+3+y+2}{y^4}$

$=\frac{3y^2+7y+5}{y^4}$

$=\frac{3y^2}{y^4}+\frac{7y}{y^4}+\frac{5}{y^4}$

$=\frac{3}{y^2}+\frac{7}{y^3}+\frac{5}{y^4}$

$=\frac{3}{{(x-1)}^2}+\frac{7}{{(x-1)}^3}+\frac{5}{{(x-1)}^4}$