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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (2 a x - x^{2})

x = \frac{( y - 3 ) ^{2}}{5}, x = 13 - y

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

\int_{0}^{8} \frac{9}{10 x ^{2} + 11 x + 1} d x

\frac{d y}{d x} = \frac{x ^{2}}{6 e ^{y}}

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