`f(x) = x^3.4 - 2x^(sqrt(2) - 1)` Find the most general antiderivative of the function. (Check your answer by differentiation.)

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Chapter 4, 4.9 - Problem 8 - 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 (x^(3.4) - 2x^(sqrt2 - 1))dx = int (x^(3.4))dx - int (2x^(sqrt2 - 1))dx`

You need to use the following formula:

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

`int (x^(3.4))dx = (x^(3.4+1))/(3.4+1) + c = (x^(4.4))/(4.4) + c`

`int (2x^(sqrt2 - 1))dx = 2*(x^(sqrt2 - 1+1))/(sqrt2 - 1+1) + c`

`int (2x^(sqrt2 - 1))dx = sqrt2*(x^(sqrt2)) + c`

Gathering all the results yields:

`int (x^(3.4) - 2x^(sqrt2 - 1))dx =(x^(4.4))/(4.4) + sqrt2*(x^(sqrt2)) + c`

Hence, evaluating the most general antiderivative of the function yields `F(x) = (x^(4.4))/(4.4) + sqrt2*(x^(sqrt2)) + c.`

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