What is the integral of 3x^2(x^3-4)^8 ?

The integral of 3x^2(x^3-4)^8 is 1/9 (x^3-4)^9+C

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integral of 3x^2(x^3-4)^8

\int 3 x^{2} (x^{3} - 4)^{8} d x

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

Solution steps

\int 3 x^{2} (x^{3} - 4)^{8} d x

Take the constant out:

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

= 3 \cdot \int x^{2} (x^{3} - 4)^{8} d x

Apply u-substitution

= 3 \cdot \int \frac{u ^{8}}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int u^{8} d u

Apply the Power Rule

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{9}}{9}

Substitute back

u = x^{3} - 4

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of 3x^2(x^3-4)^8 ?
The integral of 3x^2(x^3-4)^8 is 1/9 (x^3-4)^9+C