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$If f^{'} (x) = \tan^{- 1} (\sec x + \tan x), - \frac{π}{2} < x < \frac{π}{2}, and f (0) = 0, then f (1) is equal to:$

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π+24

π−14

π+14

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Correct option is D

f'(x)=tan-1(secx+tanx) =tan-11+sinxcosx=tan-11-cosπ2+xsinπ2+x=tan-12sin2π4+x22sinπ4+x2cosπ4+x2 =tan-1tanπ4+x2=π4+x2 ∫f'(x)dx=∫π4+x2dx f(x)=π4x+x24+cf(0)=c=0⇒f(x)=π4x+x24 So f(1)=π+14

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If f'(x)=tan-1(secx+tanx),-π2<x<π2, and f(0)=0, then f(1) is equal to:

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$If f^{'} (x) = \tan^{- 1} (\sec x + \tan x), - \frac{π}{2} < x < \frac{π}{2}, and f (0) = 0, then f (1) is equal to:$