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

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

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

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

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

Solution steps

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

Expand

(2 x - 1) (x^{2} + 2) : 2 x^{3} + 4 x - x^{2} - 2

= \int 2 x^{3} + 4 x - x^{2} - 2 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 x^{3} d x + \int 4 x d x - \int x^{2} d x - \int 2 d x

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 4 x d x = 2 x^{2}

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

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