`f(x) = 3sqrt(x) - 2root(3)(x)` Find the most general antiderivative of the function. (Check your answer by differentiation.)

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Chapter 4, 4.9 - Problem 11 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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The most general antiderivative F(x) of the function f(x) can be found using the following relation:

`int f(x)dx = F(x) + c`

`int (3sqrt x - 2root(3) x)dx = int (3sqrt x)dx - int (2root(3) x)dx`

You need to use the following formulas:

`sqrt x = x^(1/2)`

`root(3) x = x^(1/3)`

`int x^n dx = (x^(n+1))/(n+1) `

`int (3sqrt x)dx= 3 int (x^(1/2))dx = 3*(x^(1/2+1))/(1/2+1) + c = 3*(2/3)*x^(3/2) + c`

`int (3sqrt x)dx=2x*sqrt x + c`

`int (2root(3) x)dx = 2int x^(1/3) dx = 2*(x^(1/3+1))/(1/3+1) + c = 2*(3/4)*x^(4/3) + c`

`int (2root(3) x)dx = (3/2)*x*root(3) x + c`

Gathering all the results yields:

`int (3sqrt x - 2root(3) x)dx = 2x*sqrt x - (3/2)*x*root(3) x + c`

Hence, evaluating the most general antiderivative of the function yields `F(x) = 2x*sqrt x - (3/2)*x*root(3) x + c.`

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