What is the integral of (y-2)/((y-1)(y+3)) ?

The integral of (y-2)/((y-1)(y+3)) is -1/4 ln|y-1|+5/4 ln|y+3|+C

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integral of (y-2)/((y-1)(y+3))

\int \frac{y - 2}{( y - 1 ) ( y + 3 )} d y

- \frac{1}{4} ln ∣ y - 1 ∣ + \frac{5}{4} ln ∣ y + 3 ∣ + C

Solution steps

\int \frac{y - 2}{( y - 1 ) ( y + 3 )} d y

Take the partial fraction of

\frac{y - 2}{( y - 1 ) ( y + 3 )} : - \frac{1}{4 ( y - 1 )} + \frac{5}{4 ( y + 3 )}

= \int - \frac{1}{4 ( y - 1 )} + \frac{5}{4 ( y + 3 )} 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}{4 ( y - 1 )} d y + \int \frac{5}{4 ( y + 3 )} d y

\int \frac{1}{4 ( y - 1 )} d y = \frac{1}{4} ln ∣ y - 1 ∣

\int \frac{5}{4 ( y + 3 )} d y = \frac{5}{4} ln ∣ y + 3 ∣

= - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{5}{4} ln ∣ y + 3 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{5}{4} ln ∣ y + 3 ∣ + C

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What is the integral of (y-2)/((y-1)(y+3)) ?
The integral of (y-2)/((y-1)(y+3)) is -1/4 ln|y-1|+5/4 ln|y+3|+C