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

The integral of (1-x)(1-2x)(1-3x) is x-3x^2+(11x^3)/3-(3x^4)/2+C

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

\int (1 - x) (1 - 2 x) (1 - 3 x) d x

x - 3 x^{2} + \frac{11 x ^{3}}{3} - \frac{3 x ^{4}}{2} + C

Solution steps

\int (1 - x) (1 - 2 x) (1 - 3 x) d x

Expand

(1 - x) (1 - 2 x) (1 - 3 x) : 1 - 6 x + 11 x^{2} - 6 x^{3}

= \int 1 - 6 x + 11 x^{2} - 6 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 6 x d x + \int 11 x^{2} d x - \int 6 x^{3} d x

\int 1 d x = x

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

\int 11 x^{2} d x = \frac{11 x ^{3}}{3}

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

= x - 3 x^{2} + \frac{11 x ^{3}}{3} - \frac{3 x ^{4}}{2}

Add a constant to the solution

= x - 3 x^{2} + \frac{11 x ^{3}}{3} - \frac{3 x ^{4}}{2} + C

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\int_{1}^{2} \frac{( 2 + 2 x - x ^{2} )}{8} d x

x^{^{''}} - 12 x^{^{'}} + 36 x = 2 t e^{6 t}

n = 1 \sum \infty \frac{5 ^{n}}{n ^{4}}

\int \frac{x ^{8}}{x ^{9} - 2} d x

What is the integral of (1-x)(1-2x)(1-3x) ?
The integral of (1-x)(1-2x)(1-3x) is x-3x^2+(11x^3)/3-(3x^4)/2+C