What is the integral of 4x(x-2)^3 ?

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

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integral of 4x(x-2)^3

\int 4 x (x - 2)^{3} d x

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

Solution steps

\int 4 x (x - 2)^{3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x (x - 2)^{3} d x

Expand

x (x - 2)^{3} : x^{4} - 6 x^{3} + 12 x^{2} - 8 x

= 4 \cdot \int x^{4} - 6 x^{3} + 12 x^{2} - 8 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

= 4 (\int x^{4} d x - \int 6 x^{3} d x + \int 12 x^{2} d x - \int 8 x d x)

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

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

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

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{n}{6 ^{n}}

x \to 0 lim (\frac{( 1 + x ) ^{9} - 1}{x})

\frac{\partial}{\partial x} (\frac{42}{( 5 + x ^{2} + y ^{2} )})

\int \frac{1}{1 + cos ( 2 x )} d x

\int x (x^{2} + 1)^{5} d x

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