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

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

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integral of 3(x^3-x)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int x^{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

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

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

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

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

Add a constant to the solution

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

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

\int_{1}^{10} 4 d x

y = x - 3

\frac{d}{d x} (arctan (\frac{x - 1}{x + 1}))

\frac{\partial}{\partial y} (x^{2} y - 2 x^{2} - 4 y^{2})

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