If $\int \frac{{(x-1)}^2}{{(x^2+1)}^2}dx={\tan }^{-1}x+g\left(x\right)+c$ then $g\left(x\right)=$

$\int \frac{{(x-1)}^2}{{(x^2+1)}^2}dx$

$=\int \frac{x^2+1-2x}{{(x^2+1)}^2}dx$

$=\int \left[\frac{1}{x^2+1}-\frac{2x}{{(x^2+1)}^2}\right]dx$

$={\tan }^{-1}x-\int \frac{2x}{{(x^2+1)}^2}dx , put x^2+1=t\Rightarrow 2xdx=dt$

$={\tan }^{-1}x-\int \frac{dt}{t^2}$

$={\tan }^{-1}x+\frac{1}{t}+c$

$={\tan }^{-1}x+\frac{1}{x^2+1}+c$

$\therefore g\left(x\right)=\frac{1}{x^2+1}$