<s>[INST] Maclaurin series expansion for $\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt {t}}\log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$ 
Show that the expansion of
$$\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt{t}} \log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$$
as a Maclaurin series in power of $t$ to obtain 
  $$\dfrac{4}{3}-4\sum_{n=1}^{\infty}\dfrac{t^n}{(4n^2-1)(2n+3)}$$


I am not sure how to do this. Below are some steps I have managed to get which gets more complicated and am not sure how to continue. Help is appreciated. 
Expand into: 
$$\dfrac{1}{2t}+\dfrac{1}{2} - \dfrac{(1-t)^2}{4t\sqrt{t}}\log\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)$$
$$=\dfrac{1}{2t}+\dfrac{1}{2}- \dfrac{\log\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)}{4t\sqrt{t}}(-t+1)^2$$
Multiply by congugate: 
$$=\dfrac{1}{2t}+\dfrac{1}{2}-\dfrac{\log\left(\dfrac{(1+\sqrt{t})^2}{-t+1}\right)}{4t\sqrt{t}}(-t+1)^2$$
 [/INST] Considering $$A=\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt {t}}\log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$$ just to make life easier, start making $t=x^2$ to get
$$A=\frac{2 x(1+x^2)-\left(1-x^2\right)^2 \log
   \left(\frac{1+x}{1-x}\right)}{4 x^3}$$ and use first, as @metamorphy commented, the expansion of $\log\left(\frac{1+x}{1-x}\right)$ and expand. You should arrive at
$$\frac{4}{3}-\frac{4 x^2}{15}-\frac{4 x^4}{105}-\frac{4 x^6}{315}-\frac{4
   x^8}{693}-\frac{4 x^{10}}{1287}+O\left(x^{12}\right)$$ Back to $t$
$$A=\frac 43-4\left(\frac{ t}{15}+\frac{t^2}{105}+\frac{t^3}{315}+\frac{
   t^4}{693}+\frac{t^5}{1287} \right)++O\left(t^{6}\right)$$ The denominators seem to be  what you need.
I must confess that I did not find a way to prove it without using a truncated series and finding a pattern for the coefficients.
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