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

The integral of (x-1)(x^2-2x)^2 is 1/6 (x^2-2x)^3+C

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

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

\frac{1}{6} (x^{2} - 2 x)^{3} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 2 x

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

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} - 2 x ) ^{3}}{3} : \frac{1}{6} (x^{2} - 2 x)^{3}

= \frac{1}{6} (x^{2} - 2 x)^{3}

Add a constant to the solution

= \frac{1}{6} (x^{2} - 2 x)^{3} + C

(x^{2} + 4)^{- 3}

\frac{\partial}{\partial y} (2 x^{3} + x^{2} y + 3 x y^{2})

y = x^{2} + 3

\int \frac{7 x}{e ^{x^{2}}} d x

\int \frac{1}{x ^{3}} + \frac{6}{x ^{7}} d x

What is the integral of (x-1)(x^2-2x)^2 ?
The integral of (x-1)(x^2-2x)^2 is 1/6 (x^2-2x)^3+C