What is the integral of 4(3x-2)^3 ?

The integral of 4(3x-2)^3 is 1/3 (3x-2)^4+C

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integral of 4(3x-2)^3

\int 4 (3 x - 2)^{3} d x

\frac{1}{3} (3 x - 2)^{4} + C

Solution steps

\int 4 (3 x - 2)^{3} d x

Take the constant out:

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

= 4 \cdot \int (3 x - 2)^{3} d x

Apply u-substitution

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

= 4 \cdot \int \frac{u ^{3}}{3} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 4 \cdot \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 x - 2

= 4 \cdot \frac{1}{3} \cdot \frac{( 3 x - 2 ) ^{4}}{4}

Simplify

4 \cdot \frac{1}{3} \cdot \frac{( 3 x - 2 ) ^{4}}{4} : \frac{1}{3} (3 x - 2)^{4}

= \frac{1}{3} (3 x - 2)^{4}

Add a constant to the solution

= \frac{1}{3} (3 x - 2)^{4} + C

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