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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

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