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

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

integral of y/(1-y)

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

1 - y - ln ∣ 1 - y ∣ + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Expand

\frac{- u + 1}{u} : - 1 + \frac{1}{u}

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

Apply the Sum Rule:

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

= - (- \int 1 d u + \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - (- u + ln ∣ u ∣)

Substitute back

u = 1 - y

= - (- (1 - y) + ln ∣ 1 - y ∣)

Simplify

= 1 - y - ln ∣ 1 - y ∣

Add a constant to the solution

= 1 - y - ln ∣ 1 - y ∣ + C

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