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

The integral of 1/((3x-1)^3) is -1/(6(3x-1)^2)+C

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

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

- \frac{1}{6 ( 3 x - 1 ) ^{2}} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

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

Substitute back

u = 3 x - 1

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

Simplify

\frac{1}{3} (- \frac{1}{2 ( 3 x - 1 ) ^{2}}) : - \frac{1}{6 ( 3 x - 1 ) ^{2}}

= - \frac{1}{6 ( 3 x - 1 ) ^{2}}

Add a constant to the solution

= - \frac{1}{6 ( 3 x - 1 ) ^{2}} + C

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

\frac{d}{d x} ((a x^{2} + b x + c) e^{\frac{- x}{8}})

\int \frac{x ^{2} - 2 x + 3}{x ^{3} + 3 x} d x

\frac{\partial}{\partial x} (\frac{1}{y ^{2}})

x d x + y e^{- x} d y = 0, y (0) = 1

What is the integral of 1/((3x-1)^3) ?
The integral of 1/((3x-1)^3) is -1/(6(3x-1)^2)+C