If ∫ ( x - 1 ) 2 x 2 + 1 2 d x = tan - 1 x + g x + k , then g ( x ) is equal to

Given that

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

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

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

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

We know that $\int f^{\text{'}}\left(x\right){\left[f\left(x\right)\right]}^n dx =\frac{{ f\left(x\right)}^{n+1}}{n+1}+c$

$= {\tan }^{-1}x +{\left(x^2+1\right)}^{-1}+c$

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