What is the integral of x(3x^2-5) ?

The integral of x(3x^2-5) is 1/12 (3x^2-5)^2+C

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integral of x(3x^2-5)

\int x (3 x^{2} - 5) d x

\frac{1}{12} (3 x^{2} - 5)^{2} + C

Solution steps

\int x (3 x^{2} - 5) 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} - 5

= \frac{1}{6} \cdot \frac{( 3 x ^{2} - 5 ) ^{2}}{2}

Simplify

\frac{1}{6} \cdot \frac{( 3 x ^{2} - 5 ) ^{2}}{2} : \frac{1}{12} (3 x^{2} - 5)^{2}

= \frac{1}{12} (3 x^{2} - 5)^{2}

Add a constant to the solution

= \frac{1}{12} (3 x^{2} - 5)^{2} + C

3 x^{2} - 2 y^{3} \frac{d y}{d x} = 0

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

\int x^{2} + 4 x + 3 d x

\int \frac{sec ( x )}{1 - tan ( x )} d x

\frac{d}{d x} ((x^{3} - 7) (2 x^{2} + 3))

What is the integral of x(3x^2-5) ?
The integral of x(3x^2-5) is 1/12 (3x^2-5)^2+C