What is the integral of 1/((2x-1)^3) ?

The integral of 1/((2x-1)^3) is -1/(4(2x-1)^2)+C

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integral of 1/((2x-1)^3)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= \int \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Substitute back

u = 2 x - 1

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

Simplify

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

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

Add a constant to the solution

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

x \to - 6 lim (\frac{x ^{2} + 6}{x - 6})

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\frac{d}{d x} (sin^{- 3} (x))

\frac{d}{d x} ((1 + 8 x^{2}) (x - x^{2}))

What is the integral of 1/((2x-1)^3) ?
The integral of 1/((2x-1)^3) is -1/(4(2x-1)^2)+C