If $\frac{x}{(x-1){(x^2+1)}^2}=\frac{1}{4(x-1)}-\frac{x+1}{x^2+1}+k$, then $k=$

$\frac{x}{(x-1){(x^2+1)}^2}=\frac{1}{4(x-1)}-\frac{x+1}{x^2+1}+k$

$\frac{x}{(x-1){(x^2+1)}^2}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{{(x^2+1)}^2}$

$x=A{(x^2+1)}^2+(Bx+C)(x^2+1)(x-1)+(Dx+E)(x-1)$

$x=A(x^4+1+2x^2)+Bx(x^3-x^2+x-1)+C(x^3-x^2+x-1)+Dx(x-1)+E(x-1)$

Put $x=1,4A=1\Rightarrow A=\frac{1}{4}$

Comparing $x^4$coefficients on both sides,

$A+B=0\Rightarrow B=\frac{-1}{4}$

Comparing $x^3$coefficients on both sides

$-B+C=0\Rightarrow C=-\frac{1}{4}$

Comparing $x^2$coefficients on both sides

$2A+B-C+D=0$

$\frac{2}{4}-\frac{1}{4}+\frac{1}{4}+D=0\Rightarrow D=-\frac{1}{2}$

Comparing $x$coefficients on both sides

$-B+C-D+E=1$

$\frac{1}{4}-\frac{1}{4}+\frac{1}{2}+E=1\Rightarrow E=\frac{1}{2}$

$\therefore$$\frac{x}{(x-1){(x^2+1)}^2}=\frac{1}{4(x-1)}-\frac{x+1}{x^2+1}+\frac{1-x}{2{(x^2+1)}^2}$