How do you find the integral of $((x) \sqrt{x - 1}) d x$ $((x)sqrt(x-1))dx$ ?

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Answer 1 · 2015-09-19T00:53:22

$(\frac{2}{5}) {(x - 1)}^{\frac{5}{2}} + (\frac{2}{3}) {(x - 1)}^{\frac{3}{2}} + C$ $(2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + C$

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

that is dx = du

after substitution integral changes to

Integral ((u+1) $\sqrt{u}$ $sqrt(u)$ ) du

= $\int (u^{\frac{3}{2}} + u^{\frac{1}{2}}) d u$ $int (u^(3/2) + u^(1/2))du$

= $\frac{u^{\frac{5}{2}}}{\frac{5}{2}} + \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C$ $u^(5/2)/(5/2) + u^(3/2)/(3/2) + C$

= $(\frac{2}{5}) u^{\frac{5}{2}} + (\frac{2}{3}) u^{\frac{3}{2}} + C$ $(2/5)u^(5/2) + (2/3)u^(3/2) + C$

= $(\frac{2}{5}) {(x - 1)}^{\frac{5}{2}} + (\frac{2}{3}) {(x - 1)}^{\frac{3}{2}} + C$ $(2/5)(x-1)^(5/2) + (2/3)(x-1)^(3/2) + C$

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

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