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

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