What is the integral of (1-x)(6+x^2) ?

The integral of (1-x)(6+x^2) is 6x+(x^3)/3-3x^2-(x^4)/4+C

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integral of (1-x)(6+x^2)

\int (1 - x) (6 + x^{2}) d x

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

Solution steps

\int (1 - x) (6 + x^{2}) d x

Expand

(1 - x) (6 + x^{2}) : 6 + x^{2} - 6 x - x^{3}

= \int 6 + x^{2} - 6 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 6 d x + \int x^{2} d x - \int 6 x d x - \int x^{3} d x

\int 6 d x = 6 x

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

\int 6 x d x = 3 x^{2}

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

= 6 x + \frac{x ^{3}}{3} - 3 x^{2} - \frac{x ^{4}}{4}

Add a constant to the solution

= 6 x + \frac{x ^{3}}{3} - 3 x^{2} - \frac{x ^{4}}{4} + C

sin (x \frac{π}{2})

\int e^{x} \cdot x^{3} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}

\int ln (x) \frac{1}{x} d x

What is the integral of (1-x)(6+x^2) ?
The integral of (1-x)(6+x^2) is 6x+(x^3)/3-3x^2-(x^4)/4+C