What is the integral of (6x-1)^3 ?

The integral of (6x-1)^3 is 1/24 (6x-1)^4+C

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integral of (6x-1)^3

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

\frac{1}{24} (6 x - 1)^{4} + C

Solution steps

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

Apply u-substitution

= \int \frac{u ^{3}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{3} d u

Apply the Power Rule

= \frac{1}{6} \cdot \frac{u ^{4}}{4}

Substitute back

u = 6 x - 1

= \frac{1}{6} \cdot \frac{( 6 x - 1 ) ^{4}}{4}

Simplify

\frac{1}{6} \cdot \frac{( 6 x - 1 ) ^{4}}{4} : \frac{1}{24} (6 x - 1)^{4}

= \frac{1}{24} (6 x - 1)^{4}

Add a constant to the solution

= \frac{1}{24} (6 x - 1)^{4} + C

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