What is the integral of (x^3-3x+2) ?

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

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integral of (x^3-3x+2)

\int (x^{3} - 3 x + 2) d x

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

Solution steps

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

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

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

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