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

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

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integral of 1/((y^2-y))

integral

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

- ln ∣ y ∣ + ln ∣ y - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{y ^{2} - y} : - \frac{1}{y} + \frac{1}{y - 1}

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

\int \frac{1}{y} d y = ln ∣ y ∣

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

= - ln ∣ y ∣ + ln ∣ y - 1 ∣

Add a constant to the solution

= - ln ∣ y ∣ + ln ∣ y - 1 ∣ + C

\frac{1}{a - x}

(e^{x})^{- 1}

\frac{1}{x ( x + 1 )}

\frac{( x ^{2} + 2 x - 1 )}{( x ^{3} - x )}

ln (\frac{1}{x})

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