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 Evaluate 1x21+x2dx

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a
−x2+1x+c
b
−xx2+1+c
c
−x2-1x+c
d
None of these

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detailed solution

Correct option is A

Let x = tanθ or dx = sec2θ dθ ∴ ∫1x21+x2dx=∫sec2⁡θdθtan2⁡θsec⁡θ                                      =∫cosec⁡θcot⁡θdθ                                      =−cosec⁡θ+c                                      =−x2+1x+c


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(1cosx)cosec2xdx is equal to


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