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Question: If I = <img src="https://cdn.pureessence.tech/canvas_509.png?top_left_x=300&top_left_y=0&width=300&h...

If I = , then f(p) is equal to

A
B

(12cos⁡−1p+12sec⁡−1p)\left( \frac { 1 } { 2 } \cos ^ { - 1 } p + \frac { 1 } { 2 } \sec ^ { - 1 } p \right)

C

ln⁡p+p2−1−12sec⁡−1p\ln \sqrt { p + \sqrt { p ^ { 2 } - 1 } } - \frac { 1 } { 2 } \sec ^ { - 1 } p

D

None of these

Answer

ln⁡p+p2−1−12sec⁡−1p\ln \sqrt { p + \sqrt { p ^ { 2 } - 1 } } - \frac { 1 } { 2 } \sec ^ { - 1 } p

Explanation

Solution

Let I = ∫12pp−1p+1dp\int \frac { 1 } { 2 p } \sqrt { \frac { p - 1 } { p + 1 } } d p I

= 12∫p−1p(p+1)(p−1)dp\frac { 1 } { 2 } \int \frac { p - 1 } { p \sqrt { ( p + 1 ) ( p - 1 ) } } d p =

= - 12sec⁡−1p\frac { 1 } { 2 } \sec ^ { - 1 } p