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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{2} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = x^{2} - 2 x

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

Simplify

\frac{1}{2} \cdot \frac{2}{3} (x^{2} - 2 x)^{\frac{3}{2}} : \frac{1}{3} (x^{2} - 2 x)^{\frac{3}{2}}

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

Add a constant to the solution

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

\frac{d}{d h} (35 f (h) (3 + 2 h))

(1 + \frac{x}{2})^{- 4}

\int 16 - x d x

3 x^{2} y^{^{''}} + 7 x y^{^{'}} + y = 24 x + 2

\frac{d}{d x} (3 x^{2} - 4 x y^{4} - y^{3})

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