How do you integrate $int (sqrtx-1)^2/sqrtx$ using substitution?

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Answer 1 · 2016-11-22T06:01:36

The answer is $=(2(sqrtx-1)^3)/3+C$

Use the substitution

$sqrtx=u$ , $=>$ , $x=u^2$

$dx=2udu=2sqrtxdu$

$int((sqrtx-1)^2dx)/sqrtx$

$=int(u-1)^2*(2cancelsqrtxdu)/cancelsqrtx$

$=2int(u-1)^2du$

$=(2(u-1)^3)/3$

$=(2(sqrtx-1)^3)/3+C$

How do you integrate int (sqrtx-1)^2/sqrtxint (sqrtx-1)^2/sqrtx using substitution?