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

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

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

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

ln ∣ y ∣ - ln ∣ 1 - y ∣ + C

Solution steps

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

\int \frac{1}{y} d y = ln ∣ y ∣

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

= ln ∣ y ∣ - ln ∣ 1 - y ∣

Add a constant to the solution

= ln ∣ y ∣ - ln ∣ 1 - y ∣ + C

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\int \frac{x ^{2} - 5 x - 24}{x + 3} d x

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