What is the integral of x(x-1)^{-1/2} ?

The integral of x(x-1)^{-1/2} is 2/3 (x-1)^{3/2}+2(x-1)^{1/2}+C

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integral of x(x-1)^{-1/2}

\int x (x - 1)^{- \frac{1}{2}} d x

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

Solution steps

\int x (x - 1)^{- \frac{1}{2}} d x

Apply u-substitution

: \int \frac{u + 1}{u ^{\frac{1}{2}}} d u

= \int \frac{u + 1}{u ^{\frac{1}{2}}} d u

Expand

\frac{u + 1}{u ^{\frac{1}{2}}} : u^{\frac{1}{2}} + \frac{1}{u ^{\frac{1}{2}}}

= \int u^{\frac{1}{2}} + \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{\frac{1}{2}} d u + \int \frac{1}{u ^{\frac{1}{2}}} d u

\int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}}

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

= \frac{2}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}}

Substitute back

u = x - 1

= \frac{2}{3} (x - 1)^{\frac{3}{2}} + 2 (x - 1)^{\frac{1}{2}}

Add a constant to the solution

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

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What is the integral of x(x-1)^{-1/2} ?
The integral of x(x-1)^{-1/2} is 2/3 (x-1)^{3/2}+2(x-1)^{1/2}+C