What is the integral of 9(3x-6)^{-5} ?

The integral of 9(3x-6)^{-5} is -1/(108(x-2)^4)+C

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integral of 9(3x-6)^{-5}

\int 9 (3 x - 6)^{- 5} d x

- \frac{1}{108 ( x - 2 ) ^{4}} + C

Solution steps

\int 9 (3 x - 6)^{- 5} d x

Take the constant out:

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

= 9 \cdot \int (3 x - 6)^{- 5} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= 9 \cdot \int \frac{1}{( 3 x - 6 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{5}} d u

= 9 \cdot \int \frac{1}{3 u ^{5}} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= 9 \cdot \frac{1}{3} (- \frac{1}{4 u ^{4}})

Substitute back

u = 3 x - 6

= 9 \cdot \frac{1}{3} (- \frac{1}{4 ( 3 x - 6 ) ^{4}})

Simplify

9 \cdot \frac{1}{3} (- \frac{1}{4 ( 3 x - 6 ) ^{4}}) : - \frac{1}{108 ( x - 2 ) ^{4}}

= - \frac{1}{108 ( x - 2 ) ^{4}}

Add a constant to the solution

= - \frac{1}{108 ( x - 2 ) ^{4}} + C

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What is the integral of 9(3x-6)^{-5} ?
The integral of 9(3x-6)^{-5} is -1/(108(x-2)^4)+C