What is the integral of 1/((1-y)^2) ?

The integral of 1/((1-y)^2) is 1/(1-y)+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)^2)

\int \frac{1}{( 1 - y ) ^{2}} d y

\frac{1}{1 - y} + C

Solution steps

\int \frac{1}{( 1 - y ) ^{2}} d y

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= - (- \frac{1}{u})

Substitute back

u = 1 - y

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

Simplify

= \frac{1}{1 - y}

Add a constant to the solution

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