What is the integral of (3x-1)^{20} ?

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

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integral of (3x-1)^{20}

\int (3 x - 1)^{20} d x

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

Solution steps

\int (3 x - 1)^{20} d x

Apply u-substitution

: \int \frac{u ^{20}}{3} d u

= \int \frac{u ^{20}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{20} d u

Apply the Power Rule

: \frac{u ^{21}}{21}

= \frac{1}{3} \cdot \frac{u ^{21}}{21}

Substitute back

u = 3 x - 1

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

Simplify

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

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

Add a constant to the solution

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

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