What is the integral of x^2(6x^3-2)^3 ?

The integral of x^2(6x^3-2)^3 is 1/72 (6x^3-2)^4+C

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integral of x^2(6x^3-2)^3

\int x^{2} (6 x^{3} - 2)^{3} d x

\frac{1}{72} (6 x^{3} - 2)^{4} + C

Solution steps

\int x^{2} (6 x^{3} - 2)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{18} d u

= \int \frac{u ^{3}}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{18} \cdot \frac{u ^{4}}{4}

Substitute back

u = 6 x^{3} - 2

= \frac{1}{18} \cdot \frac{( 6 x ^{3} - 2 ) ^{4}}{4}

Simplify

\frac{1}{18} \cdot \frac{( 6 x ^{3} - 2 ) ^{4}}{4} : \frac{1}{72} (6 x^{3} - 2)^{4}

= \frac{1}{72} (6 x^{3} - 2)^{4}

Add a constant to the solution

= \frac{1}{72} (6 x^{3} - 2)^{4} + C

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What is the integral of x^2(6x^3-2)^3 ?
The integral of x^2(6x^3-2)^3 is 1/72 (6x^3-2)^4+C