`f(t) = (3t^4 - t^3 + 6t^2)/(t^4)` Find the most general antiderivative of the function. (Check your answer by differentiation.)

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Chapter 4, 4.9 - Problem 14 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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The most general antiderivative F(t) of the function f(t) can be found using the following relation:

`int f(t)dt = F(t) + c`

`int (3t^4 - t^3 + 6t^2)/(t^4)dt = int (3t^4)/(t^4)dt - int (t^3)/(t^4)dt + int 6t^2/(t^4) dt`

You need to use the following formula:

`int t^n dt = (t^(n+1))/(n+1) `

`int (3t^4)/(t^4)dt = int 3dt = 3t + c`

`int (t^3)/(t^4)dt = int (1/t) dt = ln |t| + c`

`int 6t^2/(t^4) dt = 6 int 1/(t^2) dt = -6/t + c`

Gathering all the results yields:

`int (3t^4 - t^3 + 6t^2)/(t^4)dt = 3t - ln |t| -6/t + c`

Hence, evaluating the most general antiderivative of the function yields `F(t) = 3t - ln |t| -6/t + c.`

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