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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{ln ∣ y + 1 ∣}{2} - \frac{ln ∣ y - 1 ∣}{2} + C

Solution steps

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

Expand

(1 - y) (1 + y) : 1 - y^{2}

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

Use the common integral:

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

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

Add a constant to the solution

= \frac{ln ∣ y + 1 ∣}{2} - \frac{ln ∣ y - 1 ∣}{2} + C

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