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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

y^{^{'}} = 3 + t - y

\frac{d}{d x} (10 e^{- x})

\frac{\partial}{\partial x} (ln (4 x^{2}))

\frac{d ^{2}}{d x ^{2}} (e^{x} sin (x))

\int e^{\frac{t}{50}} d t

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