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

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

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integral of 1/((y-3))

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

ln ∣ y - 3 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = y - 3

= ln ∣ y - 3 ∣

Add a constant to the solution

= ln ∣ y - 3 ∣ + C

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