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

The integral of 1/((3x-1)^6) is -1/(15(3x-1)^5)+C

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integral of 1/((3x-1)^6)

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

- \frac{1}{15 ( 3 x - 1 ) ^{5}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u ^{6}} d u

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

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= \frac{1}{3} (- \frac{1}{5 u ^{5}})

Substitute back

u = 3 x - 1

= \frac{1}{3} (- \frac{1}{5 ( 3 x - 1 ) ^{5}})

Simplify

\frac{1}{3} (- \frac{1}{5 ( 3 x - 1 ) ^{5}}) : - \frac{1}{15 ( 3 x - 1 ) ^{5}}

= - \frac{1}{15 ( 3 x - 1 ) ^{5}}

Add a constant to the solution

= - \frac{1}{15 ( 3 x - 1 ) ^{5}} + C

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What is the integral of 1/((3x-1)^6) ?
The integral of 1/((3x-1)^6) is -1/(15(3x-1)^5)+C