`f'(t) = 4/(1 + t^2), f(1) = 0` Find `f`.

Textbook Question

Chapter 4, 4.9 - Problem 33 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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The general antiderivative of f' is

`f(t)=4arctan(t)+C,`

where C is any constant.

It must be f(1)=0, so C=-4arctan(1). acrctan(1) is `pi/4,` so the answer is

`f(t)=4arctan(t)-pi.`

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