How do you integrate #int (2-sqrtx)^5/sqrtx# using substitution?

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Answer 1 · 2017-06-24T22:56:05

#int (2-sqrtx)^5/sqrtx dx = - (2-sqrtx)^6/3+C#

Substitute:

#t = 2-sqrtx#

#dt = -dx/(2sqrtx)#

thus:

#int (2-sqrtx)^5/sqrtx dx = -2 int t^5dt =-t^6/3 + C#

and undoing the substitution:

#int (2-sqrtx)^5/sqrtx dx = - (2-sqrtx)^6/3+C#