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If f'(x) = tan⁻¹(sec x + tan x), -π/2 < x < π/2, and f(0) = 0, then f(1) is equal to

1. (π - 1)/4
2. 1/4
3. (π + 2)/4
4. (π + 1)/4

Correct answer: (π + 1)/4

Solution

Since sec x + tan x = tan(pi/4 + x/2), f'(x)=arctan(tan(pi/4 + x/2)) = pi/4 + x/2. Integrating: f(x)=pi*x/4 + x^2/4 (f(0)=0). Then f(1)=pi/4 + 1/4 = (pi+1)/4.

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