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

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

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integral of 3/(y-3)

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

3 ln ∣ y - 3 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 ln ∣ u ∣

Substitute back

u = y - 3

= 3 ln ∣ y - 3 ∣

Add a constant to the solution

= 3 ln ∣ y - 3 ∣ + C

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