What is the integral of x(4-2x^2) ?

The integral of x(4-2x^2) is -1/8 (4-2x^2)^2+C

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integral of x(4-2x^2)

\int x (4 - 2 x^{2}) d x

- \frac{1}{8} (4 - 2 x^{2})^{2} + C

Solution steps

\int x (4 - 2 x^{2}) d x

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{4} \cdot \frac{u ^{2}}{2}

Substitute back

u = 4 - 2 x^{2}

= - \frac{1}{4} \cdot \frac{( 4 - 2 x ^{2} ) ^{2}}{2}

Simplify

- \frac{1}{4} \cdot \frac{( 4 - 2 x ^{2} ) ^{2}}{2} : - \frac{1}{8} (4 - 2 x^{2})^{2}

= - \frac{1}{8} (4 - 2 x^{2})^{2}

Add a constant to the solution

= - \frac{1}{8} (4 - 2 x^{2})^{2} + C

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What is the integral of x(4-2x^2) ?
The integral of x(4-2x^2) is -1/8 (4-2x^2)^2+C