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

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

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

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

- ln ∣ y - 1 ∣ + ln ∣ y - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

f (s) = 2 t e^{- 5 t}

f (x) = x^{2} + 2 x, [1, 2]

\frac{d}{d x} (- 2 x^{2} + 2 x)

x \to 1 - lim (1 - e^{\frac{1}{x - 1}})

\int \frac{1}{17 - x ^{2}} d x

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