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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 2 x - 1

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

Simplify

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

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

Add a constant to the solution

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

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

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\int_{- \infty}^{- 1} 0 d x

\int \frac{18 x - 6}{3 x ^{2} - 2 x - 7} d x

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