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

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

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

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

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

Solution steps

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

Expand

(x - 1)^{2} x^{3} (2 - x) : - x^{6} + 4 x^{5} - 5 x^{4} + 2 x^{3}

= \int - x^{6} + 4 x^{5} - 5 x^{4} + 2 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 4 x^{5} d x - \int 5 x^{4} d x + \int 2 x^{3} d x

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

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

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

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{5 x}{9 y}

\frac{d}{d x} (7 e^{e})

\frac{\partial}{\partial x} ((x^{3} + y^{2})^{\frac{1}{2}})

y^{^{''}} + 2 y^{^{'}} + 2 y = 1

\frac{d}{d x} (\frac{7 - sec ( x )}{tan ( x )})

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