<s>[INST] New series formula for $\arctan(x)$? Ln(x)? I discovered this equation, but have no idea if it has been previously discovered.  Please help determine if it has been previously developed. Or please prove that the equation is not correct.
$$\sum_{n=0}^\infty \frac{x^{2n+1}}{(x^2+1)^{n+1}}\cdot\frac{(2n)!!}{(2n+1)!!}=\arctan(x),$$
for $|x|\leq \pi$, or possibly all $x$.
Likewise, using the same method
for $x> .001$, or possibly  x > 0.
$$\sum_{n=1}^\infty \frac{x^{n}-1}{(1+x)^{n}}\cdot\frac{(1)}{(n)}=Ln(x),$$
all follows from dx/dx =1.
 [/INST] Evidently from marty cohen's answer, the formula for $\arctan$ was previously known, and as shown below, the result for $\log(x)$ is not too hard to derive. That you have discovered them yourself is no less impressive however.

Substituting $x\mapsto4x^2$ in $(2)$ from this answer, we get
$$
\sum_{k=0}^\infty\frac{4^kx^{2k}}{\binom{2k}{k}}=\frac1{1-x^2}\left[1+\frac{x}{\sqrt{1-x^2}}\sin^{-1}(x)\right]\tag{1}
$$
Integrating $(1)$ gives
$$
\begin{align}
\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}}
&=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\\
&=\frac{\tan^{-1}\left(\frac{\large x}{\sqrt{1-x^2}}\right)}{\sqrt{1-x^2}}\tag{2}
\end{align}
$$
Substituting $x\mapsto\frac{x}{\sqrt{1+x^2}}$ in $(2)$ yields
$$
\frac1{\sqrt{1+x^2}}\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}\left(1+x^2\right)^k}
=\sqrt{1+x^2}\tan^{-1}(x)\tag{3}
$$
and therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}\left(1+x^2\right)^{k+1}}
=\tan^{-1}(x)}\tag{4}
$$

$$
\begin{align}
\sum_{n=1}^\infty\frac{x^n-1}{n(1+x)^n}
&=\log\left(1-\frac1{1+x}\right)-\log\left(1-\frac{x}{1+x}\right)\\
&=\log(x)\tag{5}
\end{align}
$$
</s>