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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(3-y)

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

- ln ∣ 3 - y ∣ + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 3 - y

= - ln ∣ 3 - y ∣

Add a constant to the solution

= - ln ∣ 3 - y ∣ + C