Answer
$\Sigma_{n=1}^{\infty} (-1)^{n+1} \dfrac{x^{2n}}{n}$
Work Step by Step
Here, we have: $f(x)=\ln (x^2+1) \implies f(0)=0\\ f'(x)=\dfrac{2x}{x^2+1} \implies f'(0)=0\\ f''(x)=\dfrac{2-2x^2}{(x^2+1)^2}\implies f''0)=2$
and so on.
The Taylor series of a function $f(x)$ at $x=c$ can be written as:
$T(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+.......$
Hence, the required Taylor series is:
$T(x)=0+0(x-0)+\dfrac{2}{2!}(x-0)^2+0(x-0)^3+....=x^2-\dfrac{x^4}{2}+\dfrac{x^6}{3}+....=\Sigma_{n=1}^{\infty} (-1)^{n+1} \dfrac{x^{2n}}{n}$
