What is the integral of (1-2x^2+3x^3) ?

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

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integral of (1-2x^2+3x^3)

\int (1 - 2 x^{2} + 3 x^{3}) d x

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

Solution steps

\int 1 - 2 x^{2} + 3 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 1 d x - \int 2 x^{2} d x + \int 3 x^{3} d x

\int 1 d x = x

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

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

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

Add a constant to the solution

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

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