What is the integral of 6x^2(2x^3-1)^{99} ?

The integral of 6x^2(2x^3-1)^{99} is 1/100 (2x^3-1)^{100}+C

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integral of 6x^2(2x^3-1)^{99}

\int 6 x^{2} (2 x^{3} - 1)^{99} d x

\frac{1}{100} (2 x^{3} - 1)^{100} + C

Solution steps

\int 6 x^{2} (2 x^{3} - 1)^{99} 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} - 1)^{99} d x

Apply u-substitution

: \int \frac{u ^{99}}{6} d u

= 6 \cdot \int \frac{u ^{99}}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int u^{99} d u

Apply the Power Rule

: \frac{u ^{100}}{100}

= 6 \cdot \frac{1}{6} \cdot \frac{u ^{100}}{100}

Substitute back

u = 2 x^{3} - 1

= 6 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} - 1 ) ^{100}}{100}

Simplify

6 \cdot \frac{1}{6} \cdot \frac{( 2 x ^{3} - 1 ) ^{100}}{100} : \frac{1}{100} (2 x^{3} - 1)^{100}

= \frac{1}{100} (2 x^{3} - 1)^{100}

Add a constant to the solution

= \frac{1}{100} (2 x^{3} - 1)^{100} + C

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What is the integral of 6x^2(2x^3-1)^{99} ?
The integral of 6x^2(2x^3-1)^{99} is 1/100 (2x^3-1)^{100}+C