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

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

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

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

\frac{1}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Simplify

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

= \int \frac{1}{2 x - 1} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 1

= \frac{1}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

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