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

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

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

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

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

Solution steps

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

Simplify

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

= \int x - 1 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 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

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

\int sin^{5} (\frac{x}{2}) cos^{4} (\frac{x}{2}) d x

\frac{\partial}{\partial x} (y + x y)

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

\frac{d}{d x} (x + y + z^{2})

\frac{\partial}{\partial x} (- 2 k l (ln (\frac{y}{x - y})))

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