Question

find the value of then I think u have a idea​

Answer 1

Answer:

Is my answer is true I hope it helps u

Answer 2

Answer:

$\frac{ {x}^{2} }{2} - 2x + ln |x| + c$

Step-by-step explanation:

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