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

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

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integral of (x^3+2x^2-3x+4)

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

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

Solution steps

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

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\int x e^{- \frac{x}{6}} d x

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

\int y^{2} e^{- 2 y} d y

f (x) = x e^{\frac{1}{x}}

- cos (x), [0, π]

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