`g(t) = (1 + t + t^2)/(sqrt(t))` Find the most general antiderivative of the function.

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Chapter 4, 4.9 - Problem 15 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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The most general antiderivative G(t) of the function g(t) can be found using the following relation:

`int g(t)dt = G(t) + c`

`int (1 + t + t^2)/(sqrt t)dt = int (1)/(sqrt t)dt + int (t)/(sqrt t)dt + int t^2/(sqrt t) dt`

You need to use the following formula:

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

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

`int (t)/(sqrt t)dt= int (sqrt t*sqrt t)/(sqrt t)dt = int sqrt t dt = (t^(1/2+1))/(1/2+1) + c = (2/3)*t^(3/2) + c = (2/3)*tsqrt t + c`

`int t^2/(sqrt t) dt = int t^(2-1/2)dt = int t^(3/2)dt = (t^(3/2+1))/(3/2+1) + c`

`int t^2/(sqrt t) dt = (2/5)*t^(5/2) + c = (2/5)*t^2*sqrt t + c`

Gathering all the results yields:

`int (1 + t + t^2)/(sqrt t)dt = 2t^(1/2) + (2/3)*tsqrt t + (2/5)*t^2*sqrt t + c`

Hence, evaluating the most general antiderivative of the function yields `G(t) = 2t^(1/2) + (2/3)*tsqrt t + (2/5)*t^2*sqrt t + c.`

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