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Question: \(\int \frac { d x } { ( 1 + \sqrt { x } ) \sqrt { \left( x - x ^ { 2 } \right) } }\) is equal to...

∫dx(1+x)(x−x2)\int \frac { d x } { ( 1 + \sqrt { x } ) \sqrt { \left( x - x ^ { 2 } \right) } } is equal to

A

1+x(1−x)2+c\frac { 1 + \sqrt { x } } { ( 1 - x ) ^ { 2 } } + c

B

1+x(1+x)2+c\frac { 1 + \sqrt { x } } { ( 1 + x ) ^ { 2 } } + c

C

1−x(1−x)2+c\frac { 1 - \sqrt { x } } { ( 1 - x ) ^ { 2 } } + c

D

2(x−1)(1−x)+c\frac { 2 ( \sqrt { x } - 1 ) } { \sqrt { ( 1 - x ) } } + c

Answer

2(x−1)(1−x)+c\frac { 2 ( \sqrt { x } - 1 ) } { \sqrt { ( 1 - x ) } } + c

Explanation

Solution

Let I = ∫dx(1+x)(x−x2)\int \frac { d x } { ( 1 + \sqrt { x } ) \sqrt { \left( x - x ^ { 2 } \right) } }

If x=sin⁡p\sqrt { x } = \sin p, then 12xdx=cos⁡pdp\frac { 1 } { 2 \sqrt { x } } d x = \cos p d p ⇒ I

= = = 2

∫(1−sin⁡p)dpcos⁡2p\int \frac { ( 1 - \sin p ) d p } { \cos ^ { 2 } p }

= 2

= 2 (tan p - sec p) = 2(x(1−x)−1(1−x))+c2 \left( \sqrt { \frac { x } { ( 1 - x ) } } - \frac { 1 } { \sqrt { ( 1 - x ) } } \right) + c

= 2(x−1)(1−x)+c\frac { 2 ( \sqrt { x } - 1 ) } { \sqrt { ( 1 - x ) } } + c