What is the integral of (3x^2-9)^{20}2x ?

The integral of (3x^2-9)^{20}2x is 1/63 (3x^2-9)^{21}+C

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integral of (3x^2-9)^{20}2x

\int (3 x^{2} - 9)^{20} 2 x d x

\frac{1}{63} (3 x^{2} - 9)^{21} + C

Solution steps

\int (3 x^{2} - 9)^{20} \cdot 2 x d x

Take the constant out:

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

= 2 \cdot \int (3 x^{2} - 9)^{20} x d x

Apply u-substitution

= 2 \cdot \int \frac{u ^{20}}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int u^{20} d u

Apply the Power Rule

= 2 \cdot \frac{1}{6} \cdot \frac{u ^{21}}{21}

Substitute back

u = 3 x^{2} - 9

= 2 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} - 9 ) ^{21}}{21}

Simplify

2 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{2} - 9 ) ^{21}}{21} : \frac{1}{63} (3 x^{2} - 9)^{21}

= \frac{1}{63} (3 x^{2} - 9)^{21}

Add a constant to the solution

= \frac{1}{63} (3 x^{2} - 9)^{21} + C

What is the integral of (3x^2-9)^{20}2x ?
The integral of (3x^2-9)^{20}2x is 1/63 (3x^2-9)^{21}+C