#### Answer

tTe Maclaurin series for $f(x)=\ln(1-x^2)$ is convergent for the values $|-x^2|=x^2\lt 1$ and diverges at $x=\pm1$.

#### Work Step by Step

By making use of Table 2, we have the Maclaurin series for $f(x)=e^x$ as follows
$$f(x)=\ln(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$
Now by the comparison with the function $f(x)=\ln(1-x)$, we have the Maclaurin series as follows
$$f(x)=\ln(1-x^2)=-x^2-\frac{x^4}{2}-\frac{x^6}{3}-\frac{x^8}{4}+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=\ln(1+x)$ is convergent for the values $|x|\lt 1$ and $x=1$. In our case, $x=1$ leads to the log of zero, which is undefined. Hence the Maclaurin series for $f(x)=\ln(1-x^2)$ is convergent for the values $|-x^2|=x^2\lt 1$ and diverges at $x=\pm1$.