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

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

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integral of ((sqrt(x-1))^2)/(sqrt(x))

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

\frac{2}{3} x^{\frac{3}{2}} - 2 x + C

Solution steps

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

(x - 1)^{2} = x - 1

= \int \frac{x - 1}{x} d x

Expand

\frac{x - 1}{x} : x - \frac{1}{x}

= \int x - \frac{1}{x} d x

Apply the Sum Rule:

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

= \int x d x - \int \frac{1}{x} d x

\int x d x = \frac{2}{3} x^{\frac{3}{2}}

\int \frac{1}{x} d x = 2 x

= \frac{2}{3} x^{\frac{3}{2}} - 2 x

Add a constant to the solution

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

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