# 2F1(1/2,1;3/2;z^2)=log((1+z)/(1-z))/(2z)

The following formula holds: $${}_2F_1 \left( \dfrac{1}{2}, 1 ; \dfrac{3}{2}; z^2 \right)= \dfrac{1}{2z} \log \left( \dfrac{1+z}{1-z} \right),$$ where ${}_2F_1$ denotes the hypergeometric 2F1 and $\log$ denotes the logarithm.