Answer
$x^2-\dfrac{1}{2}x^4+\dfrac{1}{3}x^{6}-\dfrac{1}{4}x^{8}+..$
Work Step by Step
Consider the Taylor's Series for $\ln (1+x)$ as follows:
$ \ln (1+x)=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{n+1}}{(n+1)!}=x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+...$
Plug $x^2$ instead of $x$ in the above series as follows:
Then $ \ln (1+x^2)=x^2-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^3}{3!}-\dfrac{(x^2)^4}{4!}+...$
$ \implies x^2-\dfrac{1}{2}x^4+\dfrac{1}{3}x^{6}-\dfrac{1}{4}x^{8}+..$
