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a
1+x(1−x)2+c
b
1+x(1+x)2+c
c
1−x(1−x)2+c
d
2(x−1)(1−x)+c
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Correct option is D
Let I=∫dx(1+x)x−x2 If x=sinp, then 12xdx=cospdpI=∫2sinpcospdp(1+sinp)sinp⋅cosp=2∫dp(1+sinp)=2∫(1−sinp)dpcos2p=2∫sec2pdp−∫(tanpsecp)dp=2(tanp−secp)=2x(1−x)−1(1−x))+C=2(x−1)(1−x)+CSimilar Questions
is equal to
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