If $I = ∫\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp=f(p)+c$, then f(p) is equal to

$\frac{1}{2}In(p-\sqrt{p^2-1})$ $(\frac{1}{2}cos^{-1}p+\frac{1}{2}sec^{-1}p)$ $In\sqrt{p+\sqrt{p^2-1}}-\frac{1}{2}sec^{-1}p$ none of these

$In\sqrt{p+\sqrt{p^2-1}}-\frac{1}{2}sec^{-1}p$

Let $I = ∫\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp$ $=\frac{1}{2}∫\frac{p-1}{p\sqrt{(p+1)(p-1)}}dp=\frac{1}{2}∫\frac{pdp}{p\sqrt{p^2-1}}-\frac{1}{2}∫\frac{dp}{p\sqrt{p^2-1}}$ $=\frac{1}{2}log_e(p+\sqrt{p^2-1})-\frac{1}{2}sec^{-1}p$. Hence (C) is the correct answer.