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

The integral of x(3x^2-2) is 1/12 (3x^2-2)^2+C

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integral of x(3x^2-2)

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

\frac{1}{12} (3 x^{2} - 2)^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{6} d u

= \int \frac{u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{6} \cdot \frac{u ^{2}}{2}

Substitute back

u = 3 x^{2} - 2

= \frac{1}{6} \cdot \frac{( 3 x ^{2} - 2 ) ^{2}}{2}

Simplify

\frac{1}{6} \cdot \frac{( 3 x ^{2} - 2 ) ^{2}}{2} : \frac{1}{12} (3 x^{2} - 2)^{2}

= \frac{1}{12} (3 x^{2} - 2)^{2}

Add a constant to the solution

= \frac{1}{12} (3 x^{2} - 2)^{2} + C

\frac{d}{d x} (\frac{1}{5} x^{5} - 3 x^{4} + 9 x^{2} + 1)

\frac{\partial}{\partial x} (cos (2 x + y) - z)

\frac{d y}{d x} = \frac{6 y}{x}

x \to 0 lim (\frac{sin ( 6 x )}{2})

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

What is the integral of x(3x^2-2) ?
The integral of x(3x^2-2) is 1/12 (3x^2-2)^2+C