How do you find the integral of #((x)sqrt(x-1))dx#?

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Answer 1 · 2015-09-19T00:53:22

#(2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + C #

Let x-1 = u
this gives x = u+1

that is dx = du

after substitution integral changes to

Integral ((u+1) #sqrt(u)#) du

= #int (u^(3/2) + u^(1/2))du#

= #u^(5/2)/(5/2) + u^(3/2)/(3/2) + C#

= #(2/5)u^(5/2) + (2/3)u^(3/2) + C#

= #(2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + C#

Answer: (2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + C