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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 6 \cdot \int \frac{( 2 u - 3 ) ^{3}}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int (2 u - 3)^{3} d u

Apply u-substitution

: \int \frac{v ^{3}}{2} d v

= 6 \cdot \frac{1}{3} \cdot \int \frac{v ^{3}}{2} d v

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= 6 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{v ^{4}}{4}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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