What is the integral of (3x-6)^{23} ?

The integral of (3x-6)^{23} is 1/72 (3x-6)^{24}+C

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integral of (3x-6)^{23}

\int (3 x - 6)^{23} d x

\frac{1}{72} (3 x - 6)^{24} + C

Solution steps

\int (3 x - 6)^{23} d x

Apply u-substitution

: \int \frac{u ^{23}}{3} d u

= \int \frac{u ^{23}}{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^{23} d u

Apply the Power Rule

: \frac{u ^{24}}{24}

= \frac{1}{3} \cdot \frac{u ^{24}}{24}

Substitute back

u = 3 x - 6

= \frac{1}{3} \cdot \frac{( 3 x - 6 ) ^{24}}{24}

Simplify

\frac{1}{3} \cdot \frac{( 3 x - 6 ) ^{24}}{24} : \frac{1}{72} (3 x - 6)^{24}

= \frac{1}{72} (3 x - 6)^{24}

Add a constant to the solution

= \frac{1}{72} (3 x - 6)^{24} + C

sin (3 t)

\int u^{- \frac{1}{3}} d u

\frac{d}{d x} (x x - 1)

y = \frac{x}{( x - 2 ) ^{2}}

\frac{d}{d x} ((x^{2} + 1) \cdot (x + 5 + \frac{1}{x}))

What is the integral of (3x-6)^{23} ?
The integral of (3x-6)^{23} is 1/72 (3x-6)^{24}+C