If 1x(1-x)/(1+x),dx = g(x)+c, g(1)=0, then g(12) is equal to:

Question

If $\displaystyle\int\frac1x\sqrt{\frac{1-x}{1+x}}\,dx = g(x)+c$, $g(1)=0$, then $g\left(\dfrac12\right)$ is equal to:

Accepted Answer

*(The full substitution steps were not fully recoverable from the source text; using $x=\cos	heta$ and standard log/arctan antiderivatives, with $g(1)=0$ fixing the constant, evaluation at $x=1/2$ yields the stated answer.)*

Solutions · 1