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

The integral of (y+2)/(y-1) is y-1+3ln|y-1|+C

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

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

y - 1 + 3 ln ∣ y - 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

= y - 1 + ln ∣ y - 1 ∣ + 2 ln ∣ y - 1 ∣

Simplify

= y - 1 + 3 ln ∣ y - 1 ∣

Add a constant to the solution

= y - 1 + 3 ln ∣ y - 1 ∣ + C

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What is the integral of (y+2)/(y-1) ?
The integral of (y+2)/(y-1) is y-1+3ln|y-1|+C