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

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

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integral of 2x(x^2-2)

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int x (x^{2} - 2) d x

Expand

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

= 2 \cdot \int x^{3} - 2 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

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

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

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

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

Add a constant to the solution

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

f (x) = (x + 2) 2 x - 3, at x = 2

n = 1 \sum \infty (\frac{π}{5})^{n}

y = \frac{1}{x}, y = x, x = 3

\frac{d}{d x} (- \frac{4 x}{x ^{2} + 1})

\int \frac{8 - 128 x}{9 - 64 x ^{2}} d x

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