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

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

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integral of 2x^3-6x^2

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

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

Solution steps

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

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

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

Add a constant to the solution

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

\frac{d y}{d x} + y = \frac{1 - e ^{- 2 x}}{e ^{x} + e ^{- x}}

cos (3 cos (2 u))

\frac{d}{d x} (x^{- 4} + cos (x))

f (x) = \frac{cos ( t )}{t ^{7}}

x \to - 3 lim (2 x + 1)