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

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

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integral of (x^3-2x)^2(3x^2-2)

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = x^{3} - 2 x

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

Add a constant to the solution

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

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