How do you integrate $\int \frac{{(\sqrt{x} - 1)}^{2}}{\sqrt{x}}$ $int (sqrtx-1)^2/sqrtx$ using substitution?

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Answer 1 · 2016-11-22T06:01:36

The answer is $= \frac{2 {(\sqrt{x} - 1)}^{3}}{3} + C$ $=(2(sqrtx-1)^3)/3+C$

Use the substitution

$\sqrt{x} = u$ $sqrtx=u$ , $\Rightarrow$ $=>$ , $x = u^{2}$ $x=u^2$

$d x = 2 u d u = 2 \sqrt{x} d u$ $dx=2udu=2sqrtxdu$

$\int \frac{{(\sqrt{x} - 1)}^{2} d x}{\sqrt{x}}$ $int((sqrtx-1)^2dx)/sqrtx$

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

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

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

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

How do you integrate int (sqrtx-1)^2/sqrtx∫(√x−1)2√xint (sqrtx-1)^2/sqrtx using substitution?