What is the integral of (1-x)(7+x^2) ?

The integral of (1-x)(7+x^2) is 7x+(x^3)/3-(7x^2)/2-(x^4)/4+C

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integral of (1-x)(7+x^2)

\int (1 - x) (7 + x^{2}) d x

7 x + \frac{x ^{3}}{3} - \frac{7 x ^{2}}{2} - \frac{x ^{4}}{4} + C

Solution steps

\int (1 - x) (7 + x^{2}) d x

Expand

(1 - x) (7 + x^{2}) : 7 + x^{2} - 7 x - x^{3}

= \int 7 + x^{2} - 7 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 7 d x + \int x^{2} d x - \int 7 x d x - \int x^{3} d x

\int 7 d x = 7 x

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

\int 7 x d x = \frac{7 x ^{2}}{2}

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

= 7 x + \frac{x ^{3}}{3} - \frac{7 x ^{2}}{2} - \frac{x ^{4}}{4}

Add a constant to the solution

= 7 x + \frac{x ^{3}}{3} - \frac{7 x ^{2}}{2} - \frac{x ^{4}}{4} + C

What is the integral of (1-x)(7+x^2) ?
The integral of (1-x)(7+x^2) is 7x+(x^3)/3-(7x^2)/2-(x^4)/4+C