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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ((y+1))/((y-1))

integral

\frac{( y + 1 )}{( y - 1 )}

2 ln ∣ y - 1 ∣ + y + C

Solution steps

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

\frac{y + 1}{y - 1} = \frac{2}{y - 1} + 1

= \int \frac{2}{y - 1} + 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2}{y - 1} d y + \int 1 d y

\int \frac{2}{y - 1} d y = 2 ln ∣ y - 1 ∣

\int 1 d y = y

= 2 ln ∣ y - 1 ∣ + y

Add a constant to the solution

= 2 ln ∣ y - 1 ∣ + y + C

in t e g r a l (- 1)^{x}

in t e g r a l \frac{1}{( x ^{2} )}

in t e g r a l sin (2 x + 3)

in t e g r a l x^{2} (e^{x})^{2}

in t e g r a l x \cdot x^{2}

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