Question

find the integral of the function

Answer 1

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Answer 2

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$= > \int \: {( \sqrt{x} - \frac{1}{ \sqrt{x} } )}^{2} dx \\ \\ = > \int \: (x + \frac{1}{x} - 2 \sqrt{x} . \frac{ 1 }{ \sqrt{x} } )dx \\ \\ = > \int \: (x + \frac{1}{x} - 2)dx \\ \\ = > \int xdx + \int \frac{1}{x} dx - 2 \int dx \\ \\ = > \frac{ {x}^{2} }{2} + log(x) - 2x + c$
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