If $I = ∫\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp=f(p)+c$, then f(p) is equal to
Answer & explanation
Correct answer: option 3
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.