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

The integral of 1/((2y-3)) is 1/2 ln|2y-3|+C

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integral of 1/((2y-3))

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

\frac{1}{2} ln ∣ 2 y - 3 ∣ + C

Solution steps

\int \frac{1}{2 y - 3} d y

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 y - 3

= \frac{1}{2} ln ∣ 2 y - 3 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 y - 3 ∣ + C

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What is the integral of 1/((2y-3)) ?
The integral of 1/((2y-3)) is 1/2 ln|2y-3|+C