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

The integral of (3x-6)^3 is 1/12 (3x-6)^4+C

Popular >

integral of (3x-6)^3

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

\frac{1}{12} (3 x - 6)^{4} + C

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

u = 3 x - 6

= \frac{1}{3} \cdot \frac{( 3 x - 6 ) ^{4}}{4}

Simplify

\frac{1}{3} \cdot \frac{( 3 x - 6 ) ^{4}}{4} : \frac{1}{12} (3 x - 6)^{4}

= \frac{1}{12} (3 x - 6)^{4}

Add a constant to the solution

= \frac{1}{12} (3 x - 6)^{4} + C

\int (x - 1) cos (x) d x

\int \frac{2 x + 1}{( x + 2 ) ^{2}} d x

\int \frac{4}{x x} d x

\int (\frac{1}{x ^{2} + 1})^{2} d x

\int \frac{2 x + 1}{9 - x ^{2}} d x