What is the integral of 8x^2(3x^3-1)^{16} ?

The integral of 8x^2(3x^3-1)^{16} is 8/153 (3x^3-1)^{17}+C

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integral of 8x^2(3x^3-1)^{16}

\int 8 x^{2} (3 x^{3} - 1)^{16} d x

\frac{8}{153} (3 x^{3} - 1)^{17} + C

Solution steps

\int 8 x^{2} (3 x^{3} - 1)^{16} d x

Take the constant out:

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

= 8 \cdot \int x^{2} (3 x^{3} - 1)^{16} d x

Apply u-substitution

= 8 \cdot \int \frac{u ^{16}}{9} d u

Take the constant out:

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

= 8 \cdot \frac{1}{9} \cdot \int u^{16} d u

Apply the Power Rule

= 8 \cdot \frac{1}{9} \cdot \frac{u ^{17}}{17}

Substitute back

u = 3 x^{3} - 1

= 8 \cdot \frac{1}{9} \cdot \frac{( 3 x ^{3} - 1 ) ^{17}}{17}

Simplify

8 \cdot \frac{1}{9} \cdot \frac{( 3 x ^{3} - 1 ) ^{17}}{17} : \frac{8}{153} (3 x^{3} - 1)^{17}

= \frac{8}{153} (3 x^{3} - 1)^{17}

Add a constant to the solution

= \frac{8}{153} (3 x^{3} - 1)^{17} + C

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What is the integral of 8x^2(3x^3-1)^{16} ?
The integral of 8x^2(3x^3-1)^{16} is 8/153 (3x^3-1)^{17}+C