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

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

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

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

\frac{x ^{5}}{5} - x^{4} + \frac{4 x ^{3}}{3} + C

Solution steps

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

Expand

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

= \int x^{4} - 4 x^{3} + 4 x^{2} 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^{4} d x - \int 4 x^{3} d x + \int 4 x^{2} d x

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

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

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

= \frac{x ^{5}}{5} - x^{4} + \frac{4 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{5}}{5} - x^{4} + \frac{4 x ^{3}}{3} + C

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