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

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

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

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

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

Solution steps

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

Expand

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

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

\int 2 d x = 2 x

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

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

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

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

Add a constant to the solution

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

\frac{s - 9}{s ^{2} - 9}

\frac{d}{d x} (tan (4 x))

\frac{d y}{d x} + \frac{y}{x - 2} = 5 (x - 2) y^{\frac{1}{2}}

f (x) = \frac{6 x ^{2} - 1}{2 x}

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