What is the integral of 6/((2x-1)^2) ?

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

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integral of 6/((2x-1)^2)

\int \frac{6}{( 2 x - 1 ) ^{2}} d x

- \frac{3}{2 x - 1} + C

Solution steps

\int \frac{6}{( 2 x - 1 ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{( 2 x - 1 ) ^{2}} d x

Apply u-substitution

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

= 6 \cdot \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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 6 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = 2 x - 1

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

Simplify

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

= - \frac{3}{2 x - 1}

Add a constant to the solution

= - \frac{3}{2 x - 1} + C

\frac{d}{d x} (\frac{x}{x + 20})

\int \frac{2 x}{x + 2} d x

\frac{d y}{d x} = 0.0225 y + 0.0003 y^{2}

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

\frac{\partial}{\partial x} (x^{2} y + 3 x y + y^{2})

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