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#