Answer
$$f(x)=\tan^{-1}(x^2)= x^2-\frac{x^6}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+...$$
Convergent for $|x|\leq 1$
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=\tan^{-1}x$ as follows
$$f(x)=\tan^{-1}= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...$$
Now by the comparison with the function $f(x)=\tan^{-1}(x^2)$, we have the Maclaurin series as follows
$$f(x)=\tan^{-1}(x^2)= x^2-\frac{x^6}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=\tan^{-1}x$ is convergent for the values $|x|\leq 1$. Hence the Maclaurin series for $f(x)=\tan^{-1}(x^2)$ is convergent for the values $|x^2|=x^2\leq 1$ or $|x|\leq 1$.
