Use the power series To determine a power series centered at 0, for the function.Identify the interval of convergence. f(x)=ln(x^2+1)

$g(x)=\frac{1}{1+x}=1-x+x^2-x^3+x^4$
We must recognize a connection between 1/(1+x) and the naturallog. well if we integrate what we have now we'd end up with $\ln (1+x)$, so that's a start, but we want to end up with $\ln (x^2+1)$. So we'd need $2x/[1+x^2]$. So we have
$g(x^2)=\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8$
Multiplying everything by x
$xg(x^2)=\frac{x}{1+x^2}=x-x^3+x^5-x^7+x^9$
And by 2
$2xg(x^2)=\frac{2x}{1+x^2}=2x-2x^3+2x^5-2x^7+2x^9$
Integrating letting $u=1+x^{2}du=2xdx$, we find
$\ln (1+x^2)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\frac{x^{10}}{5}$
Hence
$\ln (1+x^2)=\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{x^{2n}}{n}$
$\underset{n\to \mathrm{\infty }}{lim}|\frac{x^{2n+1}}{n+1}\ast \frac{n}{x^{2n}}|<1$
$|x|\underset{n\to \mathrm{\infty }}{lim}|\frac{n}{n+1}|<1$
|x|<1
Now we must check the endpoints individually. For x =1, we have a alternating harmonic series, which isconvergent. Same for x = -1. So the intervalis [-1,1]

$1-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+x^{10}-x^{11}+x^{12}-x^{13}+x^{14}-x^{15}+....................$
$f(x)=\ln (x^2+1)=x^2-x^{4/2}+x^{6/3}-x^{8/4}+x^{10/5}-x^{12/6}+............................$