What is the integral of x(x-6)^2 ?

The integral of x(x-6)^2 is (x^4)/4-4x^3+18x^2+C

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integral of x(x-6)^2

\int x (x - 6)^{2} d x

\frac{x ^{4}}{4} - 4 x^{3} + 18 x^{2} + C

Solution steps

\int x (x - 6)^{2} d x

Expand

x (x - 6)^{2} : x^{3} - 12 x^{2} + 36 x

= \int x^{3} - 12 x^{2} + 36 x 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^{3} d x - \int 12 x^{2} d x + \int 36 x d x

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

\int 12 x^{2} d x = 4 x^{3}

\int 36 x d x = 18 x^{2}

= \frac{x ^{4}}{4} - 4 x^{3} + 18 x^{2}

Add a constant to the solution

= \frac{x ^{4}}{4} - 4 x^{3} + 18 x^{2} + C

\int 2 e^{- \frac{x}{5}} d x

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

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

\frac{d y}{d x} = \frac{x}{4 y}

\frac{x ^{2} + 1}{x ^{2} - 1}

What is the integral of x(x-6)^2 ?
The integral of x(x-6)^2 is (x^4)/4-4x^3+18x^2+C