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

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

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

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

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

Solution steps

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

Expand

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

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

\int 1 d x = x

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

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

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

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

Add a constant to the solution

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

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\int \frac{e ^{\frac{y}{x}}}{x} d x

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