What is the integral of 6x^2(x^3-9)^3 ?

The integral of 6x^2(x^3-9)^3 is 1/2 (x^3-9)^4+C

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integral of 6x^2(x^3-9)^3

\int 6 x^{2} (x^{3} - 9)^{3} d x

\frac{1}{2} (x^{3} - 9)^{4} + C

Solution steps

\int 6 x^{2} (x^{3} - 9)^{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} (x^{3} - 9)^{3} d x

Apply u-substitution

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

= 6 \cdot \int \frac{u ^{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 u^{3} d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = x^{3} - 9

= 6 \cdot \frac{1}{3} \cdot \frac{( x ^{3} - 9 ) ^{4}}{4}

Simplify

6 \cdot \frac{1}{3} \cdot \frac{( x ^{3} - 9 ) ^{4}}{4} : \frac{1}{2} (x^{3} - 9)^{4}

= \frac{1}{2} (x^{3} - 9)^{4}

Add a constant to the solution

= \frac{1}{2} (x^{3} - 9)^{4} + C

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What is the integral of 6x^2(x^3-9)^3 ?
The integral of 6x^2(x^3-9)^3 is 1/2 (x^3-9)^4+C