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

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

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

\int \frac{1}{( y - 1 ) ( y - 2 )} d y

- ln ∣ y - 1 ∣ + ln ∣ y - 2 ∣ + C

Solution steps

\int \frac{1}{( y - 1 ) ( y - 2 )} d y

Take the partial fraction of

\frac{1}{( y - 1 ) ( y - 2 )} : - \frac{1}{y - 1} + \frac{1}{y - 2}

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

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

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

= - ln ∣ y - 1 ∣ + ln ∣ y - 2 ∣

Add a constant to the solution

= - ln ∣ y - 1 ∣ + ln ∣ y - 2 ∣ + C

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