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

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

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

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

\frac{x ^{6}}{3} - \frac{3 x ^{4}}{4} + C

Solution steps

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

Expand

x^{3} (2 x^{2} - 3) : 2 x^{5} - 3 x^{3}

= \int 2 x^{5} - 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 2 x^{5} d x - \int 3 x^{3} d x

\int 2 x^{5} d x = \frac{x ^{6}}{3}

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

= \frac{x ^{6}}{3} - \frac{3 x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{6}}{3} - \frac{3 x ^{4}}{4} + C

\frac{\partial}{\partial x} (4 x y (- 12 - x - y))

\frac{1}{s ( s - 1 )}

x \to 3 lim ((1 - 3 x)^{\frac{1}{x}})

\int tan (2 x) d x

\frac{\partial}{\partial z} (6 y (z + 1)^{4})

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