How do you integrate $\int \frac{{(2 - \sqrt{x})}^{5}}{\sqrt{x}}$ $int (2-sqrtx)^5/sqrtx$ using substitution?

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Answer 1 · 2017-06-24T22:56:05

$\int \frac{{(2 - \sqrt{x})}^{5}}{\sqrt{x}} d x = - \frac{{(2 - \sqrt{x})}^{6}}{3} + C$ $int (2-sqrtx)^5/sqrtx dx = - (2-sqrtx)^6/3+C$

Substitute:

$t = 2 - \sqrt{x}$ $t = 2-sqrtx$

$d t = - \frac{d x}{2 \sqrt{x}}$ $dt = -dx/(2sqrtx)$

thus:

$\int \frac{{(2 - \sqrt{x})}^{5}}{\sqrt{x}} d x = - 2 \int t^{5} d t = - \frac{t^{6}}{3} + C$ $int (2-sqrtx)^5/sqrtx dx = -2 int t^5dt =-t^6/3 + C$

and undoing the substitution:

$\int \frac{{(2 - \sqrt{x})}^{5}}{\sqrt{x}} d x = - \frac{{(2 - \sqrt{x})}^{6}}{3} + C$ $int (2-sqrtx)^5/sqrtx dx = - (2-sqrtx)^6/3+C$

How do you integrate int (2-sqrtx)^5/sqrtx∫(2−√x)5√xint (2-sqrtx)^5/sqrtx using substitution?