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

The integral of (x^2-2x)^3 is (x^7)/7-x^6+(12x^5)/5-2x^4+C

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

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

\frac{x ^{7}}{7} - x^{6} + \frac{12 x ^{5}}{5} - 2 x^{4} + C

Solution steps

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

Expand

(x^{2} - 2 x)^{3} : x^{6} - 6 x^{5} + 12 x^{4} - 8 x^{3}

= \int x^{6} - 6 x^{5} + 12 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^{6} d x - \int 6 x^{5} d x + \int 12 x^{4} d x - \int 8 x^{3} d x

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

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

\int 12 x^{4} d x = \frac{12 x ^{5}}{5}

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

= \frac{x ^{7}}{7} - x^{6} + \frac{12 x ^{5}}{5} - 2 x^{4}

Add a constant to the solution

= \frac{x ^{7}}{7} - x^{6} + \frac{12 x ^{5}}{5} - 2 x^{4} + C

f (t) = \frac{8}{t ^{2}}, at t = 1

\frac{d}{d x} ((2 x - 7)^{- 1} (x + 5))

y = \frac{2 x + 3}{5 x - 1}

y = - 2 (x + 1)^{2} - 4

x \to 4 + lim (\frac{3}{x ^{2} - 16})

What is the integral of (x^2-2x)^3 ?
The integral of (x^2-2x)^3 is (x^7)/7-x^6+(12x^5)/5-2x^4+C