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

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

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integral of (1-x)^3

integral

(1 - x)^{3}

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

Solution steps

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

Expand

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

= \int 1 - 3 x + 3 x^{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 1 d x - \int 3 x d x + \int 3 x^{2} d x - \int x^{3} d x

\int 1 d x = x

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

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

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

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

Add a constant to the solution

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

e^{2} x^{2}

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

\frac{cos ( \frac{1}{x} )}{x ^{3}}

\frac{1}{1 + cos ^{2} ( x )}

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

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