What is the integral of x^3-x^2-6x ?

The integral of x^3-x^2-6x is (x^4)/4-(x^3)/3-3x^2+C

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integral of x^3-x^2-6x

\int x^{3} - x^{2} - 6 x d x

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

Solution steps

\int x^{3} - x^{2} - 6 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 x^{2} d x - \int 6 x d x

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

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

\int 6 x d x = 3 x^{2}

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

Add a constant to the solution

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

f (x) = ln (x^{5})

\frac{d}{d x} (2 + 5 x)

\int_{4}^{5} x x^{2} - 16 d x

\frac{\partial}{\partial x} (\frac{1}{3} (x^{2} + 2)^{\frac{3}{2}})

x \to \infty lim (9 x^{2} + x - 3)

What is the integral of x^3-x^2-6x ?
The integral of x^3-x^2-6x is (x^4)/4-(x^3)/3-3x^2+C