$f^\prime(x)=\frac{1}{1+x^2}\text{ }f^{\prime\prime}(x)=\frac{-2x}{(1+x^2)^2)}\text{ }f^{\prime\prime\prime}(x)=\frac{-2(1+x^2)^2-(-2x)(2(1+x^2))(2x)}{(1+x^2)^4}$

$f(0) = 0\text{ }f^\prime(0)=1\text{ }f^{\prime\prime}(0)=0\text{ }f^{\prime\prime\prime}(0)=-2$

$p_3(x)=x-\frac{2}{3!}x^3$

Use the polynomial from part (a), but replace $x$ with $x^{2}$.