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

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

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

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

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

Solution steps

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

Expand

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

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

\int 2 d x = 2 x

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

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

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

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

Add a constant to the solution

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

arctan (\frac{7}{x ^{2}})

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\frac{d}{d x} (sin (π))

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