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

The integral of (4x^2-2)x is 1/16 (4x^2-2)^2+C

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integral of (4x^2-2)x

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

\frac{1}{16} (4 x^{2} - 2)^{2} + C

Solution steps

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

Apply u-substitution

= \int \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

= \frac{1}{8} \cdot \frac{u ^{2}}{2}

Substitute back

u = 4 x^{2} - 2

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

Simplify

\frac{1}{8} \cdot \frac{( 4 x ^{2} - 2 ) ^{2}}{2} : \frac{1}{16} (4 x^{2} - 2)^{2}

= \frac{1}{16} (4 x^{2} - 2)^{2}

Add a constant to the solution

= \frac{1}{16} (4 x^{2} - 2)^{2} + C

What is the integral of (4x^2-2)x ?
The integral of (4x^2-2)x is 1/16 (4x^2-2)^2+C