`f'(t) = t + 1/(t^3), t>0, f(1) = 6` Find `f`.

Textbook Question

Chapter 4, 4.9 - Problem 34 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Integrating gives

`f(t)=t^2/2-(1/2)t^(-2)+C,`

where C is any constant.

`f(1)=1/2-1/2+C=C=6,` so C=6.

The answer: `f(t)=t^2/2-1/(2t^2)+6.`

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