If \[I=\int{\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp=f(p)+c}\], then f(p) is equal to:

\[\frac{1}{2}\ell n\left[ p-\sqrt{{{p}^{2}}-1} \right]\]

\[\frac{1}{2}{{\cos }^{-1}}p+\frac{1}{2}{{\sec }^{-1}}p\]

\[\ell n\sqrt{p+\sqrt{{{p}^{2}}-1}}-\frac{1}{2}{{\sec }^{-1}}p\]

If \[I=\int{\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp=f(p)+c}\], then f(p)

1.	If \[I=\int{\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp=f(p)+c}\], then f(p) is equal to:
A.	\[\frac{1}{2}\ell n\left[ p-\sqrt{{{p}^{2}}-1} \right]\]
B.	\[\frac{1}{2}{{\cos }^{-1}}p+\frac{1}{2}{{\sec }^{-1}}p\]
C.	\[\ell n\sqrt{p+\sqrt{{{p}^{2}}-1}}-\frac{1}{2}{{\sec }^{-1}}p\]
D.	None of the above.
Answer» D. None of the above.

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If \[I=\int{\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp=f(p)+c}\], then f(p) is equal to:

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