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

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

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

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

- ln ∣ y ∣ + ln ∣ y - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

\int \frac{x}{8 x ^{2} + 6} d x

(1 + x)^{- \frac{7}{3}}

x \to 0 lim (3 e^{2 x})

\frac{d}{d x} ((x - 1)^{3})

2 cos^{2} (x)

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