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

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

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

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

- 2 ln ∣ 1 - y ∣ - y + C

Solution steps

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

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

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

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

\int 1 d y = y

= - 2 ln ∣ 1 - y ∣ - y

Add a constant to the solution

= - 2 ln ∣ 1 - y ∣ - y + C

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