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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((y-1))

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

ln ∣ y - 1 ∣ + C

Solution steps

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

= ln ∣ y - 1 ∣

Add a constant to the solution

= ln ∣ y - 1 ∣ + C

\int x^{3} (x^{4} - 9)^{4} d x

x \to 4 lim (20 - x^{2})

\frac{\partial}{\partial y} (x^{2} + y^{2} + z^{2} - 1)

6 sin (x) - 8 e^{x}

\frac{d y}{d x} = cos (x) (1 + 3 y)