What is the integral of (4x-x^2)^2 ?

The integral of (4x-x^2)^2 is (16x^3)/3-2x^4+(x^5)/5+C

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integral of (4x-x^2)^2

\int (4 x - x^{2})^{2} d x

\frac{16 x ^{3}}{3} - 2 x^{4} + \frac{x ^{5}}{5} + C

Solution steps

\int (4 x - x^{2})^{2} d x

Expand

(4 x - x^{2})^{2} : 16 x^{2} - 8 x^{3} + x^{4}

= \int 16 x^{2} - 8 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 16 x^{2} d x - \int 8 x^{3} d x + \int x^{4} d x

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

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

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

= \frac{16 x ^{3}}{3} - 2 x^{4} + \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{16 x ^{3}}{3} - 2 x^{4} + \frac{x ^{5}}{5} + C

e^{t} cos (2 t)

- 1 {\frac{15}{s ^{2} + 4 s + 13}}

y = e^{x + 1} + 1

x \frac{d y}{d x} + 2 y = \frac{1}{x ^{3}}

x^{2} + 7 x - 5

What is the integral of (4x-x^2)^2 ?
The integral of (4x-x^2)^2 is (16x^3)/3-2x^4+(x^5)/5+C