What is the integral of (x-1)(x^2-2x)^3 ?

The integral of (x-1)(x^2-2x)^3 is (x^8)/8-x^7+3x^6-4x^5+2x^4+C

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integral of (x-1)(x^2-2x)^3

\int (x - 1) (x^{2} - 2 x)^{3} d x

\frac{x ^{8}}{8} - x^{7} + 3 x^{6} - 4 x^{5} + 2 x^{4} + C

Solution steps

\int (x - 1) (x^{2} - 2 x)^{3} d x

Expand

(x - 1) (x^{2} - 2 x)^{3} : x^{7} - 7 x^{6} + 18 x^{5} - 20 x^{4} + 8 x^{3}

= \int x^{7} - 7 x^{6} + 18 x^{5} - 20 x^{4} + 8 x^{3} 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^{7} d x - \int 7 x^{6} d x + \int 18 x^{5} d x - \int 20 x^{4} d x + \int 8 x^{3} d x

\int x^{7} d x = \frac{x ^{8}}{8}

\int 7 x^{6} d x = x^{7}

\int 18 x^{5} d x = 3 x^{6}

\int 20 x^{4} d x = 4 x^{5}

\int 8 x^{3} d x = 2 x^{4}

= \frac{x ^{8}}{8} - x^{7} + 3 x^{6} - 4 x^{5} + 2 x^{4}

Add a constant to the solution

= \frac{x ^{8}}{8} - x^{7} + 3 x^{6} - 4 x^{5} + 2 x^{4} + C

\int cos^{2} (θ) sin^{2} (θ) d θ

\int \frac{3 x + 2}{( x ^{2} + 4 x + 4 ) ( x ^{2} + 1 )} d x

\int (x - a) d x

\int (sin (θ))^{2} d θ

\frac{d}{d x} (2 x^{5} - e^{x})

What is the integral of (x-1)(x^2-2x)^3 ?
The integral of (x-1)(x^2-2x)^3 is (x^8)/8-x^7+3x^6-4x^5+2x^4+C