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

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

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

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

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

Solution steps

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

Expand

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

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

\int 1 d x = x

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

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

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

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

Add a constant to the solution

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

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